Contents / Calculus / Sequences & Series

Chapter 9

Sequences & Series

Sequences, convergence tests, power series, and Taylor series.

Introduction

A sequence is an infinite list of numbers; a series is what you get when you try to add one up. The whole chapter turns on a single question: when does adding infinitely many numbers produce a finite answer, and what is it?

The question is not idle. Every time a calculator returns $\sin(0.5)$ or $e^{1.3}$ it is summing a series, stopping when the remaining terms are too small to matter. Taylor series turn every smooth function into an "infinite polynomial", which is why they are the working tool of physics, numerical analysis and probability. And the convergence tests in the middle of the chapter are the reason we can trust those sums at all.

The chapter builds in layers. Sequences come first, because a series is defined through a sequence of partial sums. Then the two families whose sums we can actually compute, geometric and telescoping. Then the tests: divergence, integral, comparison, alternating, ratio and root, each proved and each worked on series where it succeeds and on series where it fails. The last third applies everything to power series, Taylor series and their uses.

9.1Sequences and their limits

Definition 9.1 (Sequence). A sequence is a function whose domain is the positive integers. We write its values as $a_1, a_2, a_3, \ldots$ and the whole sequence as $\{a_n\}$ or $\{a_n\}_{n=1}^{\infty}$ . The value $a_n$ is the $n$ th term.

A sequence may be given by a formula, such as $a_n = n/(n+1)$ , by a recursion, such as $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$ , or by a description, such as "the $n$ th digit of $\pi$ ". The only thing that matters for calculus is what the terms do as $n$ grows.

Definition 9.2 (Limit of a sequence). A sequence $\{a_n\}$ has limit $L$ , written $\lim_{n\to\infty} a_n = L$ or $a_n \to L$ , if for every $\varepsilon > 0$ there is an integer $N$ such that

n > N \quad\Longrightarrow\quad |a_n - L| < \varepsilon.

If such an $L$ exists the sequence converges; otherwise it diverges. We write $\lim_{n\to\infty} a_n = \infty$ when for every $M$ there is an $N$ with $a_n > M$ whenever $n > N$ .

The definition is the $\varepsilon$ - $\delta$ definition of Limits & Continuity with one change: instead of moving $x$ close to a point, you move $n$ far out along the list. Name any tolerance $\varepsilon$ , and the definition promises a place $N$ in the list beyond which every term sits within $\varepsilon$ of $L$ . A sequence like $(-1)^n$ fails: its terms are $-1, 1, -1, 1, \ldots$ , and no $L$ is within $\varepsilon = 1/2$ of both $-1$ and $1$ .

Because $a_n = f(n)$ often comes from a function $f(x)$ that makes sense for real $x$ , the limits at infinity you already know transfer directly.

Theorem 9.3 (Sequence limits from function limits). If $\lim_{x\to\infty} f(x) = L$ and $a_n = f(n)$ for every integer $n$ , then $\lim_{n\to\infty} a_n = L$ .

Proof. Given $\varepsilon > 0$ , the function limit supplies a real number $X$ with $|f(x) - L| < \varepsilon$ for all $x > X$ . Take $N$ to be any integer at least $X$ ; then every integer $n > N$ satisfies $|a_n - L| = |f(n) - L| < \varepsilon$ .∎

The theorem hands sequences every tool built for functions, L'Hôpital's Rule included. The converse fails: $\sin(\pi n) = 0$ for every integer $n$ , so the sequence converges to $0$ , while $\sin(\pi x)$ has no limit as $x\to\infty$ .

Theorem 9.4 (Limit laws for sequences). If $a_n \to L$ and $b_n \to M$ then $a_n + b_n \to L + M$ , $c\,a_n \to cL$ for any constant $c$ , $a_n b_n \to LM$ , and $a_n/b_n \to L/M$ provided $M \neq 0$ . If $f$ is continuous at $L$ then $f(a_n) \to f(L)$ .

Theorem 9.5 (Squeeze Theorem for sequences). If $a_n \le b_n \le c_n$ for all $n$ beyond some $N_0$ , and $a_n \to L$ and $c_n \to L$ , then $b_n \to L$ .

Proof. Given $\varepsilon > 0$ , choose $N \ge N_0$ so large that both $|a_n - L| < \varepsilon$ and $|c_n - L| < \varepsilon$ when $n > N$ . Then $L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon$ , which is $|b_n - L| < \varepsilon$ .∎

Corollary 9.6 (Absolute values). If $|a_n| \to 0$ then $a_n \to 0$ .

Proof. Squeeze: $-|a_n| \le a_n \le |a_n|$ , and both outer sequences tend to $0$ .∎

The corollary is the standard way to handle a factor of $(-1)^n$ . It says nothing when $|a_n|$ tends to a nonzero limit; $(-1)^n$ itself has $|a_n| = 1 \to 1$ and diverges.

Example 9.7 (Limits of explicit sequences). Find the limits of $a_n = \dfrac{n}{n+1}$ , $b_n = \dfrac{\ln n}{n}$ and $c_n = \dfrac{(-1)^n}{n}$ , and decide whether $d_n = (-1)^n$ converges.

Solution. For $a_n$ , divide top and bottom by $n$ :

\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1 + 1/n} = \frac{1}{1+0} = 1.

For $b_n$ , the form is $\infty/\infty$ , so pass to the function $f(x) = (\ln x)/x$ and use L'Hôpital's Rule:

\lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = 0,

so $b_n \to 0$ by the theorem on sequence limits from function limits. For $c_n$ , $|c_n| = 1/n \to 0$ , so $c_n \to 0$ by the corollary on absolute values. For $d_n$ , the terms alternate between $-1$ and $1$ and never settle, so $\{(-1)^n\}$ diverges.

Sanity check: $a_{100} = 100/101 \approx 0.99$ , and $b_{100} = (\ln 100)/100 \approx 0.046$ is already small and falling.□

Example 9.8 (The sequence $r^n$ ). For which real numbers $r$ does the sequence $\{r^n\}$ converge, and to what?

Solution. If $r > 1$ then $r^x$ is an increasing exponential, so $r^n \to \infty$ . If $r = 1$ every term is $1$ . If $0 < r < 1$ then $r^x$ decays to $0$ , so $r^n \to 0$ . If $r = 0$ every term is $0$ . If $-1 < r < 0$ then $|r^n| = |r|^n \to 0$ by the previous case, so $r^n \to 0$ by the corollary. If $r = -1$ the sequence is $-1, 1, -1, \ldots$ , which diverges. If $r < -1$ then $|r|^n \to \infty$ with alternating signs, which also diverges.

The whole story in one line: $\{r^n\}$ converges exactly when $-1 < r \le 1$ , with limit $0$ for $|r| < 1$ and limit $1$ for $r = 1$ . This fact drives the geometric series below.□

Example 9.9 (Squeezing a factorial). Show that $\dfrac{n!}{n^n} \to 0$ .

Solution. Write out the quotient as a product of $n$ fractions:

\frac{n!}{n^n} = \frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}.

Every factor after the first is at most $1$ , so the product is at most its first factor:

0 < \frac{n!}{n^n} \le \frac{1}{n}.

Both bounds tend to $0$ , so the Squeeze Theorem gives $n!/n^n \to 0$ . Check: for $n = 5$ the value is $120/3125 \approx 0.038$ , safely below $1/5$ .□

Some sequences have no closed formula, and for those a second route to convergence is needed: one that guarantees a limit exists without naming it.

Definition 9.10 (Monotone and bounded sequences). A sequence is increasing if $a_{n+1} \ge a_n$ for all $n$ , decreasing if $a_{n+1} \le a_n$ for all $n$ , and monotone if it is either. It is bounded above if $a_n \le M$ for some $M$ and all $n$ , bounded below if $a_n \ge m$ for some $m$ and all $n$ , and bounded if both hold.

Theorem 9.11 (Monotone Sequence Theorem). Every bounded monotone sequence converges. In particular an increasing sequence bounded above converges to the least upper bound of its terms, and a decreasing sequence bounded below converges to the greatest lower bound.

Proof. Suppose $\{a_n\}$ is increasing and bounded above. The completeness axiom of the real numbers says that every nonempty set with an upper bound has a least upper bound; let $L$ be the least upper bound of the set of terms. Given $\varepsilon > 0$ , the number $L - \varepsilon$ is not an upper bound (it is smaller than the least one), so some term satisfies $a_N > L - \varepsilon$ . Since the sequence is increasing, $a_n \ge a_N > L - \varepsilon$ for every $n > N$ , while $a_n \le L$ always. So $|a_n - L| < \varepsilon$ for $n > N$ , which is $a_n \to L$ . The decreasing case is the same argument turned upside down.∎

The completeness axiom is the one ingredient here that belongs to real analysis rather than calculus; it is what distinguishes the real numbers from the rationals, where an increasing bounded sequence like $1, 1.4, 1.41, 1.414, \ldots$ can have no limit because $\sqrt 2$ is missing. Both hypotheses are needed: $n$ is monotone but unbounded, and $(-1)^n$ is bounded but not monotone, and neither converges.

Intuition. Picture a rising water level in a tank with a lid. The water can only rise, and it can never pass the lid, so it must settle at some height, even if you cannot see where. The Monotone Sequence Theorem is that picture made precise, and it is the tool for recursively defined sequences, where you can prove the terms rise and stay under a lid long before you can write a formula for them.

Example 9.12 (A recursive sequence). Let $a_1 = 2$ and $a_{n+1} = \dfrac{a_n + 6}{2}$ . Show that $\{a_n\}$ converges and find its limit.

Solution. Compute a few terms to guess the behavior: $a_1 = 2$ , $a_2 = 4$ , $a_3 = 5$ , $a_4 = 5.5$ , $a_5 = 5.75$ . The terms seem to increase toward $6$ .

Bounded above by $6$ : this holds for $a_1 = 2$ , and if $a_n < 6$ then $a_{n+1} = (a_n + 6)/2 < (6 + 6)/2 = 6$ . By induction every term is below $6$ .

Increasing: $a_{n+1} - a_n = (a_n + 6)/2 - a_n = (6 - a_n)/2$ , which is positive because $a_n < 6$ .

So the sequence is increasing and bounded above, and by the Monotone Sequence Theorem it converges to some $L$ . Now take the limit of both sides of the recursion. Since $a_{n+1} \to L$ as well,

L = \frac{L + 6}{2} \quad\Longrightarrow\quad 2L = L + 6 \quad\Longrightarrow\quad L = 6.

Check: $a_5 = 5.75$ and $a_6 = 5.875$ , closing in on $6$ by halving the gap each step.□

Pitfall. Solving $L = (L+6)/2$ is only legitimate once you know the limit exists. The recursion $a_{n+1} = 2a_n$ with $a_1 = 1$ gives the equation $L = 2L$ , whose solution $L = 0$ is nonsense: the sequence $1, 2, 4, 8, \ldots$ diverges. Prove convergence first, then solve for the limit.

Exercises for §9.1 →

9.2Series and partial sums

Adding infinitely many numbers is not an operation anyone can perform. What we can do is add the first $N$ of them, watch the running total, and ask whether it settles.

Definition 9.13 (Series, partial sums, convergence). Given a sequence $\{a_n\}$ , the expression $\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$ is a series, and its $N$ th partial sum is

s_N = \sum_{n=1}^{N} a_n = a_1 + a_2 + \cdots + a_N.

The series converges to the sum $S$ if $\lim_{N\to\infty} s_N = S$ , and we then write $\sum_{n=1}^{\infty} a_n = S$ . If the sequence of partial sums has no finite limit, the series diverges.

So a series is a sequence in disguise: the sequence $s_1, s_2, s_3, \ldots$ of partial sums. Everything about series convergence is a statement about that sequence, and every theorem about series is proved by looking at it. The symbol $\sum_{n=1}^{\infty} a_n$ does double duty: it names the series, and when the series converges it also names the number the partial sums approach.

Intuition. Think of a savings account into which you deposit $a_1$ on day one, $a_2$ on day two, and so on forever. The series converges if the balance approaches a definite ceiling: you can name a number the account will never exceed but will get arbitrarily close to. It diverges if the balance grows without limit, or if it keeps swinging (deposits alternating with withdrawals that never fade).

Example 9.14 (Reading a series from its partial sums). Suppose the partial sums of a series $\sum a_n$ are $s_N = \dfrac{2N}{3N+1}$ . Find $a_1$ , a formula for $a_n$ , and the sum of the series.

Solution. The first partial sum is the first term: $a_1 = s_1 = 2/4 = 1/2$ . Each later term is the difference of consecutive partial sums, $a_n = s_n - s_{n-1}$ :

a_n = \frac{2n}{3n+1} - \frac{2(n-1)}{3n-2} = \frac{2n(3n-2) - 2(n-1)(3n+1)}{(3n+1)(3n-2)} = \frac{2}{(3n+1)(3n-2)}.

(Check at $n=1$ : $2/(4\cdot 1) = 1/2$ , matching $a_1$ .) The sum is the limit of the partial sums,

\sum_{n=1}^{\infty} a_n = \lim_{N\to\infty} \frac{2N}{3N+1} = \frac{2}{3}.

□

Example 9.15 (A series that visibly converges). Find the partial sums of $\sum_{n=1}^{\infty} \dfrac{1}{2^n}$ and hence its sum.

Solution. Computing directly, $s_1 = 1/2$ , $s_2 = 3/4$ , $s_3 = 7/8$ , $s_4 = 15/16$ , which suggests $s_N = 1 - 1/2^N$ . To confirm it, note $s_{N+1} = s_N + 1/2^{N+1}$ , and if $s_N = 1 - 1/2^N$ then $s_{N+1} = 1 - 1/2^N + 1/2^{N+1} = 1 - 1/2^{N+1}$ ; the formula holds by induction. Then

\sum_{n=1}^{\infty} \frac{1}{2^n} = \lim_{N\to\infty}\left(1 - \frac{1}{2^N}\right) = 1.

The general machinery for this is the geometric series formula of the next section; here the point is that "sum equals one" means exactly "the partial sums approach one".□

The single most useful consequence of the definition is that the terms of a convergent series must fade.

Theorem 9.16 (Terms of a convergent series tend to zero). If $\sum a_n$ converges, then $\lim_{n\to\infty} a_n = 0$ .

Proof. Since $a_n = s_n - s_{n-1}$ and both $s_n$ and $s_{n-1}$ tend to the sum $S$ , the limit laws give $a_n \to S - S = 0$ .∎

The converse is false, and this is the most important warning in the chapter: $a_n \to 0$ does not make $\sum a_n$ converge. The harmonic series $\sum 1/n$ has terms tending to $0$ and diverges, as proved in the section on $p$ -series. What the theorem gives is a quick way to see divergence, formalized below as the Divergence Test.

Theorem 9.17 (Algebra of convergent series). If $\sum a_n$ and $\sum b_n$ converge and $c$ is a constant, then $\sum c\,a_n$ , $\sum (a_n + b_n)$ and $\sum (a_n - b_n)$ converge, and

\sum_{n=1}^{\infty} c\,a_n = c\sum_{n=1}^{\infty} a_n, \qquad \sum_{n=1}^{\infty} (a_n \pm b_n) = \sum_{n=1}^{\infty} a_n \pm \sum_{n=1}^{\infty} b_n.

Proof. The $N$ th partial sum of $\sum (a_n + b_n)$ is $s_N + t_N$ , where $s_N$ and $t_N$ are the partial sums of the two given series; apply the limit laws for sequences. The other two statements are the same.∎

Remark. Changing, adding or removing finitely many terms never affects whether a series converges, because it changes every partial sum beyond some point by the same constant. It does change the sum. So $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=5}^{\infty} a_n$ converge or diverge together, and if they converge they differ by $a_1 + a_2 + a_3 + a_4$ .

Pitfall. The algebra theorem needs both series to converge. From $\sum 1/n$ and $\sum (-1/n)$ you cannot conclude anything about $\sum 0 = 0$ term by term: the individual series diverge, so the theorem does not apply, and in general " $\infty - \infty$ " is meaningless for series just as it is for limits.

Exercises for §9.2 →

9.3Geometric series

A geometric series is one in which each term is a fixed multiple of the one before it. It is the most important series in the chapter: the one family whose sum we can always compute exactly, and the yardstick that the ratio and root tests measure everything else against.

Definition 9.18 (Geometric series). A geometric series with first term $a \neq 0$ and common ratio $r$ is

\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots.

Equivalently, written from $n = 1$ , it is $\sum_{n=1}^{\infty} ar^{n-1}$ .

Theorem 9.19 (Geometric series). The geometric series $\sum_{n=0}^{\infty} ar^n$ converges if and only if $|r| < 1$ , and then

\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}.

If $|r| \ge 1$ the series diverges.

Proof. If $r = 1$ the partial sums are $s_N = (N+1)a$ , which diverge because $a \neq 0$ . Otherwise write the partial sum and $r$ times it:

s_N = a + ar + \cdots + ar^N, \qquad r s_N = ar + ar^2 + \cdots + ar^{N+1}.

Subtracting, every term cancels except the first of $s_N$ and the last of $rs_N$ : $s_N - rs_N = a - ar^{N+1}$ , so

s_N = \frac{a(1 - r^{N+1})}{1 - r}.

By the example on the sequence $r^n$ , $r^{N+1} \to 0$ when $|r| < 1$ , giving $s_N \to a/(1-r)$ . When $|r| > 1$ or $r = -1$ the sequence $r^{N+1}$ diverges, so $s_N$ diverges too.∎

The formula is worth reading as "first term over one minus the ratio". That phrasing does not care where the index starts: $\sum_{n=3}^{\infty} 5\cdot(1/2)^n$ has first term $5/8$ and ratio $1/2$ , so it sums to $(5/8)/(1/2) = 5/4$ . Always identify the first term actually present rather than the $a$ in a formula.

Intuition. Walk toward a wall, covering half the remaining distance with each step. Your steps are $1/2, 1/4, 1/8, \ldots$ of the way, and after $N$ steps you have covered $1 - 1/2^N$ : never all of it, but as close as you like. The sum $\sum_{n=1}^{\infty} 1/2^n = 1$ says the wall is exactly one unit away. With ratio $r = 0.9$ instead, each step covers only a tenth of what remains, and the same reasoning puts the wall at $a/(1-r) = 10a$ times the first step. With $r \ge 1$ the steps never shrink and you march off to infinity.

Example 9.20 (Summing a geometric series). Compute $\sum_{n=0}^{\infty} \dfrac{(-1)^n}{4^n}$ .

Solution. Rewrite the term as $(-1/4)^n$ , so $a = 1$ and $r = -1/4$ . Since $|r| = 1/4 < 1$ the series converges, and

\sum_{n=0}^{\infty} \left(-\frac14\right)^n = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}.

The alternating signs need no special handling; the formula is valid for negative $r$ . Check: $1 - 0.25 + 0.0625 - 0.0156 \approx 0.797$ , close to $0.8$ .□

Example 9.21 (Identifying $a$ and $r$ from a written-out series). Find the sum of $5 - \dfrac{10}{3} + \dfrac{20}{9} - \dfrac{40}{27} + \cdots$ .

Solution. Each term is the previous one multiplied by $-2/3$ : $5 \cdot (-2/3) = -10/3$ , and $(-10/3)(-2/3) = 20/9$ . So $a = 5$ and $r = -2/3$ , with $|r| < 1$ , and the sum is

\frac{5}{1 - (-2/3)} = \frac{5}{5/3} = 3.

Check: the partial sums $5, 1.67, 3.89, 2.41, 3.40, \ldots$ oscillate around $3$ with shrinking swings.□

Example 9.22 (A geometric series in disguise). Does $\sum_{n=1}^{\infty} 2^{2n}\, 3^{1-n}$ converge?

Solution. Rewrite the term in the form $ar^{n-1}$ :

2^{2n}\,3^{1-n} = \frac{4^n}{3^{n-1}} = 4\cdot\frac{4^{n-1}}{3^{n-1}} = 4\left(\frac{4}{3}\right)^{n-1}.

This is geometric with $a = 4$ and $r = 4/3 > 1$ , so it diverges. The terms $4, 16/3, 64/9, \ldots$ grow, which the Divergence Test of a later section would also catch.□

Example 9.23 (A repeating decimal as a fraction). Write $2.3\overline{17} = 2.3171717\ldots$ as a ratio of integers.

Solution. Separate the non-repeating part from the repeating block:

2.3171717\ldots = 2.3 + \frac{17}{10^3} + \frac{17}{10^5} + \frac{17}{10^7} + \cdots.

After $2.3$ the terms form a geometric series with $a = 17/1000$ and $r = 1/100$ , whose sum is

\frac{17/1000}{1 - 1/100} = \frac{17/1000}{99/100} = \frac{17}{990}.

So $2.3\overline{17} = 23/10 + 17/990 = 2277/990 + 17/990 = 2294/990 = 1147/495$ . Check: $1147/495 \approx 2.31717$ .□

Example 9.24 (A geometric series with a variable). For which $x$ does $\sum_{n=0}^{\infty} x^n$ converge, and what is its sum?

Solution. This is geometric with $a = 1$ and $r = x$ , so it converges exactly when $|x| < 1$ , and then

\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}, \qquad |x| < 1.

This single identity is the seed of every power series in the chapter: substituting, differentiating and integrating it will produce series for $\ln(1+x)$ , $\arctan x$ and much else.□

Pitfall. The formula $a/(1-r)$ is only true when $|r| < 1$ . Plugging $r = 2$ into it gives $1/(1-2) = -1$ for $1 + 2 + 4 + 8 + \cdots$ , which is absurd; the series diverges and the formula says nothing about it. Check the ratio before you sum.

Exercises for §9.3 →

9.4Telescoping series

Definition 9.25 (Telescoping series). A series $\sum a_n$ is telescoping if its terms can be written as $a_n = b_n - b_{n+1}$ for some sequence $\{b_n\}$ , so that the partial sums collapse:

s_N = (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_N - b_{N+1}) = b_1 - b_{N+1}.

The series then converges if and only if $\{b_n\}$ converges, and its sum is $b_1 - \lim_{n\to\infty} b_n$ .

Telescoping series and geometric series are the two families whose sums can be computed directly from the definition. The usual way to expose the telescoping structure is partial fractions, which turn a term like $1/(n(n+1))$ into a difference of two simpler terms.

Intuition. Imagine a hiking trail marked with altitude signs $b_1, b_2, b_3, \ldots$ at successive posts. Each term $b_n - b_{n+1}$ is the drop between consecutive posts. Add up all the drops and you learn only the total descent from the first post to wherever the trail ends; everything in between cancels out. If the trail levels off at altitude $L$ , the total descent is $b_1 - L$ .

Example 9.26 (The basic telescoping series). Compute $\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}$ .

Solution. Partial fractions give $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$ , so with $b_n = 1/n$ ,

s_N = \left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1N - \frac{1}{N+1}\right) = 1 - \frac{1}{N+1}.

As $N \to \infty$ , $s_N \to 1$ , so the sum is $1$ . Check: $s_3 = 1/2 + 1/6 + 1/12 = 3/4 = 1 - 1/4$ .□

Example 9.27 (Telescoping with a gap of two). Compute $\sum_{n=1}^{\infty} \dfrac{1}{n(n+2)}$ .

Solution. Partial fractions: $\dfrac{1}{n(n+2)} = \dfrac12\left(\dfrac1n - \dfrac{1}{n+2}\right)$ . Now the cancellation skips one term, so write out enough to see which survive:

2s_N = \left(1 - \frac13\right) + \left(\frac12 - \frac14\right) + \left(\frac13 - \frac15\right) + \cdots + \left(\frac1N - \frac{1}{N+2}\right).

Every negative $-1/k$ with $3 \le k \le N$ is cancelled by a later positive $1/k$ . What survives is the two leading positives $1 + 1/2$ and the two trailing negatives $-1/(N+1) - 1/(N+2)$ :

s_N = \frac12\left(\frac32 - \frac{1}{N+1} - \frac{1}{N+2}\right) \to \frac{3}{4}.

Check: $s_2 = 1/3 + 1/8 = 11/24 \approx 0.458$ , and the formula gives $\frac12(3/2 - 1/3 - 1/4) = 11/24$ .□

Example 9.28 (A telescoping series that diverges). Show that $\sum_{n=1}^{\infty} \ln\!\left(1 + \dfrac1n\right)$ diverges even though its terms tend to $0$ .

Solution. Since $\ln(1 + 1/n) = \ln\dfrac{n+1}{n} = \ln(n+1) - \ln n$ , the partial sums telescope:

s_N = (\ln 2 - \ln 1) + (\ln 3 - \ln 2) + \cdots + (\ln(N+1) - \ln N) = \ln(N+1) - \ln 1 = \ln(N+1).

Since $\ln(N+1) \to \infty$ , the series diverges. The terms do tend to $0$ , so this is a second example, after the harmonic series, of shrinking terms whose sum still grows without bound.□

Pitfall. Telescoping needs the partial sum written out with care. Cancelling "everything" in $\sum (1/n - 1/(n+2))$ and answering $1$ is wrong: the two-step gap leaves two survivors at the front, not one. Always write $s_N$ explicitly, then take the limit.

Exercises for §9.4 →

9.5The Divergence Test

The theorem that the terms of a convergent series tend to zero, read in the contrapositive, is the fastest test in the chapter.

Theorem 9.29 (Divergence Test). If $\lim_{n\to\infty} a_n$ does not exist, or exists and is not $0$ , then $\sum a_n$ diverges.

Proof. This is the contrapositive of the theorem that the terms of a convergent series tend to zero: if the series converged, the terms would have limit $0$ .∎

Apply it first, every time. If the terms do not shrink, no other test is needed. If they do shrink, the test says nothing at all, and you move on to the tests that follow. It never proves convergence.

Intuition. A leaking bucket is being filled with cups of water. If the cups never get smaller than, say, a tenth of a litre, the bucket overflows no matter how slowly you pour. That is the test. But cups that do shrink are no guarantee: pour $1, 1/2, 1/3, 1/4, \ldots$ litres and the bucket still overflows eventually, just very slowly.

Example 9.30 (Terms that do not tend to zero). Decide whether $\sum_{n=1}^{\infty} \dfrac{n^{2}}{5n^{2} + 4}$ and $\sum_{n=1}^{\infty} \cos\!\left(\dfrac1n\right)$ converge.

Solution. For the first, divide top and bottom by $n^{2}$ :

\lim_{n\to\infty} \frac{n^2}{5n^2+4} = \lim_{n\to\infty} \frac{1}{5 + 4/n^2} = \frac15 \neq 0,

so the series diverges by the Divergence Test. For the second, $1/n \to 0$ and $\cos$ is continuous, so $\cos(1/n) \to \cos 0 = 1 \neq 0$ : the series diverges. The partial sums of the second series are roughly $N$ after $N$ terms, which the Divergence Test predicts without any computation.□

Example 9.31 (An alternating series with no limit). Does $\sum_{n=1}^{\infty} (-1)^n \dfrac{n}{n+1}$ converge?

Solution. The absolute values $n/(n+1)$ tend to $1$ , so the terms themselves approach $-1, 1, -1, 1, \ldots$ and have no limit. By the Divergence Test the series diverges. The Alternating Series Test of a later section will not rescue it: that test requires the absolute values to tend to zero, and here they do not.□

Example 9.32 (When the test is silent). What does the Divergence Test say about $\sum_{n=1}^{\infty} \dfrac1n$ and $\sum_{n=1}^{\infty} \dfrac{1}{n^{2}}$ ?

Solution. Nothing. In both cases the terms tend to $0$ , so the hypothesis of the test is not met and it draws no conclusion. As later sections show, the first series diverges and the second converges, which is exactly why $a_n \to 0$ cannot decide the question either way.□

Pitfall. "The terms go to zero, so the series converges" is the most common false statement in this subject. The Divergence Test runs in one direction only. Terms going to zero is necessary for convergence, never sufficient.

Exercises for §9.5 →

9.6The Integral Test and estimates of sums

For a series with positive, decreasing terms there is a picture that decides convergence: draw the terms as rectangles and compare their area with the area under a curve.

Theorem 9.33 (Integral Test). Let $f$ be continuous, positive and decreasing on $[1, \infty)$ , and let $a_n = f(n)$ . Then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\int_1^{\infty} f(x)\,dx$ converges. The same holds with $1$ replaced by any starting integer $N$ .

Proof. Draw the graph of $f$ over $[1, N]$ and two sets of rectangles of width $1$ . First, rectangles with heights $a_2, a_3, \ldots, a_N$ placed on $[1,2], [2,3], \ldots, [N-1,N]$ ; because $f$ is decreasing, each rectangle sits under the curve on its interval, so

a_2 + a_3 + \cdots + a_N \le \int_1^N f(x)\,dx.

Second, rectangles with heights $a_1, a_2, \ldots, a_{N-1}$ on the same intervals; each now pokes above the curve, so

\int_1^N f(x)\,dx \le a_1 + a_2 + \cdots + a_{N-1}.

If the integral converges, the first inequality gives $s_N \le a_1 + \int_1^{\infty} f(x)\,dx$ , so the partial sums are bounded above; they increase because the terms are positive, and by the Monotone Sequence Theorem the series converges. If the integral diverges, the second inequality gives $s_{N-1} \ge \int_1^N f(x)\,dx \to \infty$ , so the series diverges.∎

The three hypotheses each do a job. Positivity makes the partial sums monotone, so that boundedness is enough for convergence. Decreasing is what lets the rectangles be trapped on one side of the curve. Continuity makes the integral meaningful. The conclusion is only about convergence: the integral and the sum are related by the inequalities in the proof but are not equal.

Intuition. Suppose the terms are the monthly cost of a service whose price decreases smoothly over time, and $f$ is the price as a continuous function of time. The total you pay over the years (the sum) and the area under the price curve (the integral) differ by at most the first month's bill; neither can be finite while the other is infinite. So to learn whether your lifetime bill is finite, integrate the curve, which calculus knows how to do, rather than summing the rectangles, which it does not.

Example 9.34 (The Integral Test on two $p$ -series). Decide whether $\sum_{n=1}^{\infty} \dfrac{1}{n^{2}}$ and $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt n}$ converge.

Solution. Both $f(x) = 1/x^{2}$ and $g(x) = 1/\sqrt x$ are continuous, positive and decreasing on $[1,\infty)$ , so the Integral Test applies. For the first,

\int_1^{\infty} \frac{dx}{x^2} = \lim_{t\to\infty}\left[-\frac1x\right]_1^t = \lim_{t\to\infty}\left(1 - \frac1t\right) = 1,

which is finite, so $\sum 1/n^{2}$ converges. For the second,

\int_1^{\infty} \frac{dx}{\sqrt x} = \lim_{t\to\infty}\left[2\sqrt x\right]_1^t = \lim_{t\to\infty}\left(2\sqrt t - 2\right) = \infty,

so $\sum 1/\sqrt n$ diverges. Note that the convergent integral equals $1$ while the sum $\sum 1/n^{2} = \pi^{2}/6 \approx 1.645$ : the test decides convergence, not the value.□

Example 9.35 (A series that needs a substitution). Decide whether $\sum_{n=2}^{\infty} \dfrac{1}{n(\ln n)^2}$ converges.

Solution. Let $f(x) = 1/(x(\ln x)^2)$ , which is continuous, positive and decreasing for $x \ge 2$ (numerator constant, denominator increasing). With $u = \ln x$ , $du = dx/x$ :

\int_2^{\infty} \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^{\infty} \frac{du}{u^2} = \left[-\frac1u\right]_{\ln 2}^{\infty} = \frac{1}{\ln 2}.

The integral is finite, so the series converges. By contrast $\sum 1/(n\ln n)$ diverges: the same substitution gives $\int du/u = \ln u$ , which is unbounded. The extra power of $\ln n$ is just enough.□

Example 9.36 (Checking "decreasing" with a derivative). Decide whether $\sum_{n=1}^{\infty} \dfrac{\ln n}{n}$ converges.

Solution. Let $f(x) = (\ln x)/x$ , which is positive for $x > 1$ . It is not obviously decreasing, so differentiate: $f'(x) = (1 - \ln x)/x^{2}$ , which is negative for $x > e$ . So $f$ is decreasing on $[3, \infty)$ , and the Integral Test applies from $n = 3$ (the first two terms do not affect convergence). With $u = \ln x$ ,

\int_3^{\infty} \frac{\ln x}{x}\,dx = \int_{\ln 3}^{\infty} u\,du = \lim_{t\to\infty}\frac{u^2}{2}\Big|_{\ln 3}^{t} = \infty.

The series diverges.□

When a series converges, the same rectangles that prove the test also say how far a partial sum is from the true sum.

Theorem 9.37 (Remainder estimate for the Integral Test). Under the hypotheses of the Integral Test, if $\sum a_n$ converges with sum $S$ and $R_N = S - s_N = a_{N+1} + a_{N+2} + \cdots$ , then

\int_{N+1}^{\infty} f(x)\,dx \le R_N \le \int_N^{\infty} f(x)\,dx.

Consequently

s_N + \int_{N+1}^{\infty} f(x)\,dx \le S \le s_N + \int_N^{\infty} f(x)\,dx.

Proof. The rectangles of heights $a_{N+1}, a_{N+2}, \ldots$ placed on $[N, N+1], [N+1, N+2], \ldots$ lie under the curve, giving $R_N \le \int_N^{\infty} f$ . Placed on $[N+1, N+2], [N+2, N+3], \ldots$ they lie above it, giving $R_N \ge \int_{N+1}^{\infty} f$ . Add $s_N$ throughout for the second display.∎

Example 9.38 (How many terms for a given accuracy?). How many terms of $\sum_{n=1}^{\infty} \dfrac{1}{n^{3}}$ are needed to approximate the sum with error at most $0.0005$ ?

Solution. The remainder satisfies

R_N \le \int_N^{\infty} \frac{dx}{x^3} = \left[-\frac{1}{2x^2}\right]_N^{\infty} = \frac{1}{2N^2}.

We need $1/(2N^{2}) \le 0.0005$ , that is $N^{2} \ge 1000$ , so $N \ge 31.6$ : take $N = 32$ terms. Check: $1/(2\cdot 32^2) = 1/2048 \approx 0.00049 < 0.0005$ , while $N = 31$ gives $1/1922 \approx 0.00052$ , not quite enough.□

Example 9.39 (Sharpening an estimate with both bounds). Estimate $\sum_{n=1}^{\infty} \dfrac{1}{n^{3}}$ using $s_{10}$ and the two-sided bound.

Solution. Direct addition gives $s_{10} = 1 + 1/8 + 1/27 + \cdots + 1/1000 \approx 1.197532$ . The bounds are

\int_{11}^{\infty} \frac{dx}{x^3} = \frac{1}{2\cdot 121} \approx 0.004132, \qquad \int_{10}^{\infty} \frac{dx}{x^3} = \frac{1}{200} = 0.005,

so $1.201664 \le S \le 1.202532$ . Taking the midpoint, $S \approx 1.2021$ with error at most $0.0005$ : the same accuracy as $32$ terms, from only ten. (The true value is $1.2020569\ldots$ )□

Pitfall. The Integral Test needs decreasing terms, and it needs positive ones. It says nothing about $\sum (-1)^n/n$ , and it can be applied to $\sum (\ln n)/n$ only after checking where $(\ln x)/x$ actually starts to decrease. Also, never report the integral as the sum: $\int_1^{\infty} dx/x^{2} = 1$ but $\sum 1/n^{2} = \pi^{2}/6$ .

Exercises for §9.6 →

9.7 $p$ -series and the harmonic series

Definition 9.40 ( $p$ -series). For a constant $p > 0$ , the series

\sum_{n=1}^{\infty} \frac{1}{n^p}

is called a $p$ -series. The case $p = 1$ , namely $\sum 1/n$ , is the harmonic series.

Theorem 9.41 ( $p$ -series test). The $p$ -series $\sum_{n=1}^{\infty} 1/n^p$ converges if $p > 1$ and diverges if $p \le 1$ .

Proof. For $p \le 0$ the terms do not tend to $0$ , so the Divergence Test applies. For $p > 0$ the function $f(x) = x^{-p}$ is continuous, positive and decreasing on $[1,\infty)$ , so the Integral Test applies. If $p \neq 1$ ,

\int_1^{t} x^{-p}\,dx = \frac{t^{1-p} - 1}{1-p},

and as $t \to \infty$ the numerator tends to $-1$ when $p > 1$ (so the integral converges to $1/(p-1)$ ) and to $\infty$ when $p < 1$ . If $p = 1$ , $\int_1^t dx/x = \ln t \to \infty$ . The series follows the integral.∎

Because the Integral Test was itself proved from the Monotone Sequence Theorem, this settles the whole family from first principles. The harmonic series also has a famous direct proof, which is worth seeing because it shows how slowly the divergence happens.

Proposition 9.42 (Divergence of the harmonic series). $\sum_{n=1}^{\infty} \dfrac1n$ diverges.

Proof. Group the terms in blocks whose lengths are powers of $2$ . Each block after the first has all its terms at least as large as its last term, and there are enough of them to total at least $1/2$ :

\underbrace{\frac13 + \frac14}_{\ge\, 2\cdot\frac14 = \frac12} + \underbrace{\frac15 + \frac16 + \frac17 + \frac18}_{\ge\, 4\cdot\frac18 = \frac12} + \underbrace{\frac19 + \cdots + \frac{1}{16}}_{\ge\, 8\cdot\frac{1}{16} = \frac12} + \cdots

So $s_{2^k} \ge 1 + k/2$ , which grows without bound. The partial sums increase past every level, so the series diverges.∎

Intuition. The harmonic series is the great counterexample of the subject: terms $1, \tfrac12, \tfrac13, \ldots$ that plainly shrink to zero, summing to infinity. The grouping proof explains how: it takes twice as many terms to add the next half, so the growth is logarithmic. Concretely $H_N \approx \ln N + 0.5772$ , so reaching a partial sum of $100$ needs about $e^{99.42} \approx 1.5\times 10^{43}$ terms. It diverges, but no computer will ever watch it happen.

The $p$ -series are the reference family for every comparison in the chapter. Once you know that $p = 1$ is the exact dividing line, a glance at how fast a term decays usually predicts the answer.

Example 9.43 (Reading off $p$ ). Decide whether $\sum_{n=1}^{\infty} \dfrac{1}{n^{4/3}}$ , $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt[3]{n}}$ and $\sum_{n=1}^{\infty} \dfrac{3}{n^{1.001}}$ converge.

Solution. The first has $p = 4/3 > 1$ , so it converges. The second is $\sum n^{-1/3}$ with $p = 1/3 \le 1$ , so it diverges. The third is $3\sum n^{-1.001}$ ; the constant $3$ does not affect convergence by the algebra of series, and $p = 1.001 > 1$ , so it converges — although extremely slowly, since $\int_1^\infty x^{-1.001}dx = 1000$ .□

Example 9.44 (A series that is not quite a $p$ -series). Decide whether $\sum_{n=1}^{\infty} \dfrac{\sqrt{n}+2}{n^{2}}$ converges.

Solution. Split it using the algebra of series:

\sum_{n=1}^{\infty} \frac{\sqrt n + 2}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} + 2\sum_{n=1}^{\infty} \frac{1}{n^2}.

Both pieces are convergent $p$ -series ( $p = 3/2$ and $p = 2$ ), so the original series converges. Splitting is legitimate here because each piece converges on its own; if one had diverged the split would prove nothing.□

Example 9.45 (Where the exponent hides). Decide whether $\sum_{n=1}^{\infty} \dfrac{n^{2} + 1}{n^{4}}$ converges, and bound its sum below.

Solution. Split: $\dfrac{n^{2}+1}{n^{4}} = \dfrac{1}{n^{2}} + \dfrac{1}{n^{4}}$ . Both are convergent $p$ -series, so the sum converges, and its value is $\pi^{2}/6 + \pi^{4}/90 \approx 1.645 + 1.082 = 2.727$ . Since every term is positive and the first term is $2$ , the sum is certainly at least $2$ , consistent with that value.□

Pitfall. A $p$ -series has a constant exponent. The series $\sum 1/2^n$ is not a $p$ -series (the variable is in the exponent, so it is geometric), and $\sum 1/n^n$ is neither. Match the shape before you apply the rule.

Exercises for §9.7 →

9.8Comparison tests

If a series with positive terms is smaller than one you know converges, it converges too. That is the whole idea, and it turns the geometric and $p$ -series into a library of reference points.

Theorem 9.46 (Direct Comparison Test). Suppose $0 \le a_n \le b_n$ for all $n$ beyond some index. If $\sum b_n$ converges, then $\sum a_n$ converges. If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Proof. Assume the inequality holds for all $n$ (dropping finitely many terms changes nothing). Let $s_N$ and $t_N$ be the partial sums of $\sum a_n$ and $\sum b_n$ ; both are increasing because the terms are nonnegative, and $s_N \le t_N$ . If $\sum b_n$ converges to $T$ , then $s_N \le t_N \le T$ for every $N$ , so $\{s_N\}$ is increasing and bounded above and converges by the Monotone Sequence Theorem. The second statement is the contrapositive of the first.∎

The test needs the right direction. To prove convergence you must dominate your series by a convergent one; to prove divergence you must minorise it by a divergent one. Bounding a series above by a divergent series proves nothing at all.

Building the inequality is the work. Making the numerator larger or the denominator smaller makes a fraction larger; making the numerator smaller or the denominator larger makes it smaller.

Example 9.47 (Comparison proving convergence). Decide whether $\sum_{n=1}^{\infty} \dfrac{1}{n^{2} + 1}$ and $\sum_{n=1}^{\infty} \dfrac{\cos^{2} n}{n^{2}}$ converge.

Solution. For the first, enlarging the fraction by shrinking the denominator gives

0 < \frac{1}{n^2+1} < \frac{1}{n^2},

and $\sum 1/n^{2}$ is a convergent $p$ -series, so the series converges. For the second, $0 \le \cos^{2} n \le 1$ gives

0 \le \frac{\cos^2 n}{n^2} \le \frac{1}{n^2},

so it converges as well. In both cases the comparison series was chosen by asking what the term looks like when the small part is discarded.□

Example 9.48 (Comparison proving divergence). Decide whether $\sum_{n=2}^{\infty} \dfrac{1}{\ln n}$ converges.

Solution. For $n \ge 2$ we have $\ln n < n$ , so shrinking the denominator enlarges the fraction:

\frac{1}{\ln n} > \frac1n > 0.

The harmonic series $\sum 1/n$ diverges, and our series dominates it term by term, so $\sum 1/\ln n$ diverges by the Direct Comparison Test.□

Direct comparison fails often, because the inequality can point the wrong way. The series $\sum 1/(n^{2} - 1)$ behaves like $\sum 1/n^{2}$ but is larger than it, so the convergent comparison is unavailable. The fix is to compare growth rates rather than values.

Theorem 9.49 (Limit Comparison Test). Suppose $a_n > 0$ and $b_n > 0$ for all large $n$ and

\lim_{n\to\infty} \frac{a_n}{b_n} = L \quad\text{with } 0 < L < \infty.

Then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Proof. Let $m = L/2$ and $M = 2L$ , both positive and finite. Since $a_n/b_n \to L$ , there is an $N$ beyond which $m < a_n/b_n < M$ , that is

m\,b_n < a_n < M\,b_n.

If $\sum b_n$ converges then so does $\sum M b_n$ , and $a_n < Mb_n$ gives convergence of $\sum a_n$ by direct comparison. If $\sum b_n$ diverges then so does $\sum m b_n$ , and $a_n > m b_n$ gives divergence of $\sum a_n$ .∎

Method 9.50 (Choosing the comparison series).

Keep only the dominant power of $n$ in the numerator and in the denominator.
Simplify; the result is almost always $b_n = 1/n^p$ or $b_n = r^n$ .
Compute $\lim a_n/b_n$ and confirm it is a finite positive number.
Decide $\sum b_n$ by the $p$ -series or geometric rule; $\sum a_n$ does the same thing.

Intuition. Two runners set off around an infinite track. Direct comparison says one is always behind the other, so if the leader stops at a finite distance the follower must too. Limit comparison is weaker but more usable: it only says the two runners' speeds stay within a constant factor of one another for the whole race. That is still enough — a constant factor cannot turn a finite total distance into an infinite one.

Example 9.51 (Limit comparison, divergent case). Decide whether $\sum_{n=1}^{\infty} \dfrac{n}{n^{2}+3}$ converges.

Solution. For large $n$ the term behaves like $n/n^{2} = 1/n$ , so take $b_n = 1/n$ :

\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n}{n^2+3}\cdot\frac{n}{1} = \lim_{n\to\infty}\frac{n^2}{n^2+3} = 1,

which is finite and positive. The harmonic series diverges, so $\sum n/(n^{2}+3)$ diverges. Note that direct comparison would have been awkward here: $n/(n^{2}+3) < 1/n$ , the wrong direction for proving divergence.□

Example 9.52 (Limit comparison, convergent case). Decide whether $\sum_{n=1}^{\infty} \dfrac{2n^{2} + 3n}{\sqrt{5 + n^{7}}}$ converges.

Solution. Dominant behaviour: the numerator is like $2n^{2}$ , the denominator like $\sqrt{n^{7}} = n^{7/2}$ , so the term is like $2n^{2}/n^{7/2} = 2/n^{3/2}$ . Take $b_n = 1/n^{3/2}$ :

\frac{a_n}{b_n} = \frac{(2n^2+3n)\,n^{3/2}}{\sqrt{5+n^7}} = \frac{2 + 3/n}{\sqrt{5/n^7 + 1}} \longrightarrow 2,

after dividing numerator and denominator by $n^{7/2}$ . Since $0 < 2 < \infty$ and $\sum n^{-3/2}$ converges ( $p = 3/2 > 1$ ), the series converges.□

Example 9.53 (A comparison against a geometric series). Decide whether $\sum_{n=1}^{\infty} \dfrac{2^n}{3^n + n}$ converges.

Solution. For large $n$ the $n$ in the denominator is negligible, so compare with $b_n = (2/3)^n$ :

\frac{a_n}{b_n} = \frac{2^n}{3^n+n}\cdot\frac{3^n}{2^n} = \frac{3^n}{3^n+n} = \frac{1}{1 + n/3^n} \longrightarrow 1,

because $n/3^n \to 0$ . Since $\sum (2/3)^n$ is a convergent geometric series, the original converges. Direct comparison also works here: $2^n/(3^n+n) < (2/3)^n$ .□

Example 9.54 (When the limit is zero or infinite). Decide whether $\sum_{n=1}^{\infty} \dfrac{\ln n}{n^{3}}$ converges.

Solution. Comparing with $b_n = 1/n^{3}$ gives $a_n/b_n = \ln n \to \infty$ , which the Limit Comparison Test does not allow. Choose a slightly weaker comparison instead: $\ln n \le n^{1/2}$ for large $n$ (since $(\ln x)/\sqrt x \to 0$ ), so

0 \le \frac{\ln n}{n^3} \le \frac{n^{1/2}}{n^3} = \frac{1}{n^{5/2}}

for $n$ large. As $\sum n^{-5/2}$ converges, direct comparison gives convergence.

(The extended form of limit comparison covers this too: if $a_n/b_n \to 0$ and $\sum b_n$ converges, so does $\sum a_n$ . It is the same argument with only the upper inequality.)□

The comparison series also bounds the error of a partial sum, because whatever dominates the terms dominates the tail.

Remark (Estimating sums by comparison). If $0 \le a_n \le b_n$ for all $n$ and $\sum b_n$ converges, then the remainders satisfy

R_N = \sum_{n=N+1}^{\infty} a_n \le \sum_{n=N+1}^{\infty} b_n,

so any estimate for the tail of the comparison series (a geometric tail summed exactly, or an integral bound for a $p$ -series tail) is also an estimate for the tail of $\sum a_n$ .

Example 9.55 (Error estimate through a comparison series). How many terms of $\sum_{n=1}^{\infty} \dfrac{1}{2^n + n}$ guarantee an error below $0.001$ ?

Solution. Since $2^n + n > 2^n$ , we have $0 < \dfrac{1}{2^n+n} < \dfrac{1}{2^n}$ , and the geometric tail sums exactly:

R_N \le \sum_{n=N+1}^{\infty} \frac{1}{2^n} = \frac{1/2^{N+1}}{1 - 1/2} = \frac{1}{2^N}.

We need $1/2^N < 0.001$ , that is $2^N > 1000$ , so $N = 10$ terms suffice ( $2^{10} = 1024$ , while $2^9 = 512$ is not enough). Check: the first omitted term is $1/(2^{11}+11) \approx 0.00049$ , comfortably below the bound.□

Pitfall. Both comparison tests require positive terms. For $\sum (-1)^n/n$ they say nothing; use the Alternating Series Test, or test $\sum |a_n|$ and invoke absolute convergence.

Exercises for §9.8 →

9.9Alternating series

Definition 9.56 (Alternating series). An alternating series is one whose terms alternate in sign, written $\sum_{n=1}^{\infty} (-1)^{n-1} b_n$ or $\sum_{n=1}^{\infty} (-1)^{n} b_n$ with $b_n > 0$ .

Theorem 9.57 (Alternating Series Test). If the positive numbers $b_n$ satisfy

b_{n+1} \le b_n \text{ for all large } n \qquad\text{and}\qquad \lim_{n\to\infty} b_n = 0,

then the alternating series $\sum (-1)^{n-1} b_n$ converges.

Proof. Look at the even partial sums. Grouping in pairs,

s_{2N} = (b_1 - b_2) + (b_3 - b_4) + \cdots + (b_{2N-1} - b_{2N}),

and every bracket is $\ge 0$ because the $b_n$ decrease, so $\{s_{2N}\}$ is increasing. Grouping the other way,

s_{2N} = b_1 - (b_2 - b_3) - \cdots - (b_{2N-2} - b_{2N-1}) - b_{2N} \le b_1,

so $\{s_{2N}\}$ is bounded above by $b_1$ . By the Monotone Sequence Theorem it converges, say to $S$ . The odd partial sums satisfy $s_{2N+1} = s_{2N} + b_{2N+1} \to S + 0 = S$ . Both subsequences tend to $S$ , so $s_N \to S$ and the series converges.∎

Both hypotheses matter. Without $b_n \to 0$ the Divergence Test kills the series, as with $\sum (-1)^n n/(n+1)$ . Without the decreasing condition the test simply does not apply, and convergence may genuinely fail even when $b_n \to 0$ .

The proof does more than prove convergence: it shows the partial sums straddle the limit, alternately overshooting and undershooting, which is exactly where the error bound comes from.

Theorem 9.58 (Alternating Series Estimation Theorem). Under the hypotheses of the Alternating Series Test, if $S = \sum (-1)^{n-1}b_n$ and $s_N$ is the $N$ th partial sum, then

|S - s_N| \le b_{N+1},

and $S - s_N$ has the same sign as the first omitted term.

Proof. The remainder after $N$ terms is itself an alternating series whose first term is $\pm b_{N+1}$ :

S - s_N = \pm(b_{N+1} - b_{N+2} + b_{N+3} - \cdots).

By the bracketing in the previous proof, the quantity in parentheses lies between $0$ and $b_{N+1}$ , which gives both the bound and the sign.∎

Intuition. Think of a pendulum losing energy, swinging past the vertical by ever-smaller amounts: right by $b_1$ , back left by $b_2$ , right again by $b_3$ . Each swing crosses the resting point, so the true value is always caught between two consecutive partial sums, and the distance left to travel is never more than the next swing. That is why alternating series are the easiest of all to approximate: the error bound is the next term, with no integral or derivative to estimate.

Example 9.59 (The alternating harmonic series). Show that $\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n}$ converges.

Solution. Here $b_n = 1/n$ . It is positive, decreasing ( $1/(n+1) < 1/n$ ), and $b_n \to 0$ . Both hypotheses hold, so by the Alternating Series Test the series converges. Its sum is $\ln 2$ , as the section on building series will show. Note that $\sum |a_n| = \sum 1/n$ diverges: the convergence here relies entirely on the cancellation of signs.□

Example 9.60 (Checking "decreasing" properly). Does $\sum_{n=1}^{\infty} (-1)^n \dfrac{n^{2}}{n^{3}+1}$ converge?

Solution. Let $b_n = n^{2}/(n^{3}+1)$ . Then $b_n \to 0$ , since dividing by $n^{3}$ gives $(1/n)/(1 + 1/n^{3}) \to 0$ . For the decreasing condition, differentiate $f(x) = x^{2}/(x^{3}+1)$ :

f'(x) = \frac{2x(x^3+1) - x^2\cdot 3x^2}{(x^3+1)^2} = \frac{x(2 - x^3)}{(x^3+1)^2},

which is negative once $x^{3} > 2$ , that is for $x \ge 2$ . So $b_n$ decreases from $n = 2$ on, and the test applies (ignoring the first term, which does not affect convergence). The series converges.□

Example 9.61 (Failure of the decreasing hypothesis). Explain why the Alternating Series Test does not apply to

\frac{1}{1} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} + \cdots

Solution. Here $b_n$ is $1/n$ for odd $n$ and $1/n^{2}$ for even $n$ , and the sequence is not decreasing: $b_2 = 1/4$ while $b_3 = 1/3 > b_2$ . So the test says nothing. In fact the series diverges: the odd-indexed terms alone form $\sum 1/(2k+1)$ , which diverges like the harmonic series, while the even-indexed terms form a convergent $p$ -series, so the partial sums grow without bound. Alternating signs do not by themselves save a series.□

Example 9.62 (Error control). How many terms of $\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n^{3}}$ are needed to estimate the sum with error less than $0.001$ ? Give that estimate to three decimal places.

Solution. By the Alternating Series Estimation Theorem the error after $N$ terms is at most $b_{N+1} = 1/(N+1)^3$ . We need

\frac{1}{(N+1)^3} < 0.001 \iff (N+1)^3 > 1000 \iff N + 1 > 10,

so $N = 10$ terms suffice (and $N = 9$ gives exactly $0.001$ , not less). Adding them:

s_{10} = 1 - \frac18 + \frac{1}{27} - \frac{1}{64} + \frac{1}{125} - \frac{1}{216} + \frac{1}{343} - \frac{1}{512} + \frac{1}{729} - \frac{1}{1000} \approx 0.9011.

So $S \approx 0.901$ with error below $0.001$ . (The true value is $0.901543\ldots$ , and the next partial sum $s_{11} \approx 0.9019$ lies on the other side of it, as the sign statement of the theorem predicts.)□

Pitfall. The estimation theorem applies only when both hypotheses of the Alternating Series Test hold. If the $b_n$ are not decreasing, the first omitted term is not a bound, and for non-alternating series it is not even close — for $\sum 1/n^{2}$ the tail after $N$ terms is about $1/N$ , far bigger than the next term $1/(N+1)^2$ .

Exercises for §9.9 →

9.10Absolute and conditional convergence

The tests so far split into two camps: those for positive terms and one for alternating terms. A series with irregular signs, such as $\sum \sin n/n^{2}$ , fits neither. The way in is to strip the signs off.

Definition 9.63 (Absolute and conditional convergence). The series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Theorem 9.64 (Absolute convergence implies convergence). If $\sum |a_n|$ converges, then $\sum a_n$ converges, and $\left|\sum a_n\right| \le \sum |a_n|$ .

Proof. For every $n$ , $0 \le a_n + |a_n| \le 2|a_n|$ . Since $\sum 2|a_n|$ converges, the Direct Comparison Test makes $\sum (a_n + |a_n|)$ converge. Then

\sum a_n = \sum \big(a_n + |a_n|\big) - \sum |a_n|

converges as a difference of two convergent series, by the algebra of series. The inequality follows from $|s_N| \le \sum_{n\le N}|a_n|$ and taking limits.∎

This is what makes the positive-term tests universal: apply any of them to $\sum |a_n|$ , and a "converges" verdict transfers to the original series with its signs restored. A "diverges" verdict does not transfer — $\sum |(-1)^{n-1}/n| = \sum 1/n$ diverges, yet the alternating harmonic series converges.

Intuition. Absolute convergence means the series converges for the robust reason: the terms are small. Conditional convergence means it converges for a fragile reason: the terms are not small enough on their own, and only the cancellation between positives and negatives saves it. The difference shows up when you reorder the terms — a robust sum does not notice, a fragile one can be broken.

Remark (Rearrangement). If $\sum a_n$ converges absolutely, every rearrangement of its terms converges to the same sum. If it converges conditionally, Riemann's rearrangement theorem says the terms can be reordered to sum to any prescribed real number, or to diverge. The alternating harmonic series sums to $\ln 2$ in its usual order, but a suitable reordering sums to $\tfrac32\ln 2$ , or to $7$ , or to $-\pi$ . The proof belongs to real analysis; the idea is that the positive terms alone and the negative terms alone each add up to infinity, so you can always take as many of one sign as you need before switching.

Example 9.65 (Absolute convergence via comparison). Classify $\sum_{n=1}^{\infty} \dfrac{\sin n}{n^{2}}$ .

Solution. Take absolute values: $\left|\dfrac{\sin n}{n^{2}}\right| \le \dfrac{1}{n^{2}}$ , and $\sum 1/n^{2}$ converges. By the Direct Comparison Test $\sum |\sin n / n^{2}|$ converges, so the series converges absolutely. The erratic signs of $\sin n$ never had to be understood, only bounded.□

Example 9.66 (Telling the two kinds apart). Classify $\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{\sqrt n}$ and $\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n^{2}}$ as absolutely convergent, conditionally convergent or divergent.

Solution. First series: the absolute values give $\sum 1/\sqrt n$ , a $p$ -series with $p = 1/2 \le 1$ , which diverges, so it is not absolutely convergent. But $b_n = 1/\sqrt n$ is positive, decreasing and tends to $0$ , so the Alternating Series Test gives convergence. It converges conditionally.

Second series: $\sum 1/n^{2}$ converges ( $p = 2$ ), so this one converges absolutely.

The only structural difference is the exponent, and it lands on either side of the $p = 1$ boundary.□

Example 9.67 (Divergence detected through absolute values). Classify $\sum_{n=1}^{\infty} \dfrac{(-1)^n n}{2n+5}$ .

Solution. Do not start with absolute values here; start with the Divergence Test. The absolute values $n/(2n+5) \to 1/2 \neq 0$ , so $a_n$ has no limit and the series diverges. Neither the alternating test (which needs $b_n \to 0$ ) nor absolute convergence applies.□

Pitfall. Absolute convergence is a one-way street. Showing that $\sum |a_n|$ diverges tells you nothing about $\sum a_n$ ; you must still test the signed series, usually with the Alternating Series Test.

Exercises for §9.10 →

9.11The Ratio and Root Tests

The comparison tests need a reference series supplied by hand. The Ratio and Root Tests automate the comparison with a geometric series: they measure the eventual ratio (or the eventual $n$ th root) of the terms and compare it with $1$ .

Theorem 9.68 (Ratio Test). Let $\displaystyle L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ (allowing $L = \infty$ ).

If $L < 1$ , then $\sum a_n$ converges absolutely. If $L > 1$ (including $L = \infty$ ), then $\sum a_n$ diverges. If $L = 1$ , the test is inconclusive.

Proof. Suppose $L < 1$ and pick $r$ with $L < r < 1$ . Since the ratios converge to $L$ , there is an $N$ beyond which $|a_{n+1}/a_n| < r$ , that is $|a_{n+1}| < r|a_n|$ . Iterating from $N$ ,

|a_{N+k}| < r^k |a_N| \quad\text{for every } k \ge 1.

So the tail $\sum_{k\ge1} |a_{N+k}|$ is dominated term by term by the convergent geometric series $|a_N|\sum r^k$ , and converges by direct comparison. Hence $\sum|a_n|$ converges and $\sum a_n$ converges absolutely.

Suppose $L > 1$ . Then beyond some $N$ we have $|a_{n+1}| > |a_n|$ , so $|a_n|$ is increasing from that point and cannot tend to $0$ . The Divergence Test applies.∎

Theorem 9.69 (Root Test). Let $\displaystyle L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$ . If $L < 1$ the series converges absolutely; if $L > 1$ it diverges; if $L = 1$ the test is inconclusive.

Proof. If $L < 1$ , choose $r$ with $L < r < 1$ . Beyond some $N$ , $\sqrt[n]{|a_n|} < r$ , so $|a_n| < r^n$ and comparison with the convergent geometric series $\sum r^n$ finishes it. If $L > 1$ then $|a_n| > 1$ for infinitely many $n$ , so the terms do not tend to $0$ .∎

Both proofs are the same manoeuvre: show the terms are eventually dominated by a geometric sequence with ratio less than $1$ . That is why both fail at $L = 1$ , the boundary where geometric comparison carries no information — and why every $p$ -series gives $L = 1$ under both tests.

Method 9.70 (Which of the two to use).

Factorials, or products of consecutive integers: use the Ratio Test, where $(n+1)!/n! = n+1$ collapses.
Whole expressions raised to the $n$ th power, such as $\left(\frac{3n+1}{2n+5}\right)^n$ : use the Root Test, where the $n$ th root removes the exponent.
Powers of a constant alone, such as $2^n$ : either works.
Rational functions of $n$ alone: neither works ( $L = 1$ ); use comparison or the Integral Test.

Intuition. The Ratio Test asks: by what factor does a term shrink compared with the one before? If that factor settles at $0.6$ , then far out the series is behaving like $\sum 0.6^n$ — geometric, ratio below $1$ , convergent — no matter what the early terms look like. If the factor settles at $1$ , the terms are shrinking by less and less, and geometric behaviour tells you nothing: that is exactly the borderland where $\sum 1/n$ and $\sum 1/n^{2}$ live, one on each side.

Example 9.71 (Ratio Test with a factorial). Decide whether $\sum_{n=1}^{\infty} \dfrac{n!}{3^n}$ converges.

Solution. Form the ratio:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{3^{n+1}}\cdot\frac{3^n}{n!} = \frac{n+1}{3}.

Then $L = \lim_{n\to\infty}(n+1)/3 = \infty > 1$ , so the series diverges. The factorial eventually outgrows any fixed exponential, and the terms blow up.□

Example 9.72 (Ratio Test, convergent case). Decide whether $\sum_{n=1}^{\infty} \dfrac{3^n}{n!}$ and $\sum_{n=1}^{\infty} \dfrac{n^{2}}{2^n}$ converge.

Solution. For the first,

\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1} \longrightarrow 0 < 1,

so it converges absolutely. For the second,

\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^n}{n^2} = \frac12\left(1 + \frac1n\right)^2 \longrightarrow \frac12 < 1,

so it converges too. Sanity check on the second: the terms $1/2, 1, 9/8, 1, 25/32, \ldots$ rise before they fall, which is fine — the test only cares about eventual behaviour.□

Example 9.73 (Root Test). Decide whether $\sum_{n=1}^{\infty} \left(\dfrac{2n+3}{3n+2}\right)^n$ converges.

Solution. The whole term is an $n$ th power, so take $n$ th roots:

\sqrt[n]{|a_n|} = \frac{2n+3}{3n+2} \longrightarrow \frac23 < 1.

By the Root Test the series converges absolutely. The Ratio Test here would require simplifying $\left(\frac{2n+5}{3n+5}\right)^{n+1}\!\big/\left(\frac{2n+3}{3n+2}\right)^{n}$ , which is far more work for the same answer.□

Example 9.74 (Root Test with an $n$ th power and a factorial-free growth). Decide whether $\sum_{n=1}^{\infty} \dfrac{n^n}{3^{1+3n}}$ converges.

Solution. Take $n$ th roots:

\sqrt[n]{|a_n|} = \frac{n}{3^{1/n}\cdot 3^{3}} = \frac{n}{27\cdot 3^{1/n}} \longrightarrow \infty,

since $3^{1/n} \to 1$ . As $L > 1$ the series diverges.□

Example 9.75 (Both tests inconclusive). What do the Ratio and Root Tests say about $\sum 1/n$ and $\sum 1/n^{2}$ ?

Solution. For $\sum 1/n$ , the ratio is $n/(n+1) \to 1$ and the root is $(1/n)^{1/n} = 1/n^{1/n} \to 1$ (because $n^{1/n}\to 1$ ). For $\sum 1/n^{2}$ , the ratio is $n^{2}/(n+1)^2 \to 1$ and the root again tends to $1$ . Both tests return $L = 1$ for both series, yet one diverges and the other converges — which is precisely what "inconclusive" means. Use the $p$ -series rule or the Integral Test instead.□

Pitfall. When the Ratio Test gives $L=1$ , do not report "diverges". It reports nothing, and you must start over with a different test. The one case where $L = 1$ does settle matters is when you can see directly that the terms fail to tend to $0$ .

Exercises for §9.11 →

9.12A strategy for testing series

With eight tests available, the skill is choosing. The order below works because each step is cheaper than the one after it.

Method 9.76 (Choosing a convergence test).

Do the terms tend to $0$ ? If not, the series diverges by the Divergence Test. This costs one limit and is never wrong.
Is it a $p$ -series $\sum 1/n^p$ or a geometric series $\sum ar^n$ ? Then quote the rule; the geometric case also gives the sum.
Does it look like a $p$ -series or geometric series with clutter attached, such as a rational function of $n$ ? Use limit comparison against the clean version.
Does it contain a factorial or an $n$ th power? Use the Ratio Test for factorials, the Root Test for $n$ th powers.
Is it alternating? Check $b_n \downarrow 0$ and use the Alternating Series Test; if you want more, test $\sum|a_n|$ for absolute convergence.
Are the signs irregular? Test $\sum |a_n|$ with a positive-term test and use absolute convergence.
Is $a_n = f(n)$ for an $f$ you can integrate, positive and decreasing? Use the Integral Test — often the last resort, and the only route for terms involving $\ln n$ .
Do consecutive terms cancel after partial fractions? It telescopes; compute the partial sum directly.

Intuition. The tests are tools with grips shaped for particular handles: factorials fit the Ratio Test, $n$ th powers fit the Root Test, logarithms fit the Integral Test, rational functions fit limit comparison, and alternating signs fit Leibniz's test. Diagnosing the shape takes a second; the computation takes a minute. Spend the second.

Example 9.77 (Diagnosing four series). For each series, name the test you would use and state the conclusion: (a) $\sum \dfrac{n^{2}-1}{3n^{2}+1}$ , (b) $\sum \dfrac{\sqrt{n^{2}+1}}{3n^{3}+2n^{2}+11}$ , (c) $\sum n^{2} e^{-n^{3}}$ , (d) $\sum \dfrac{(-1)^n}{\ln n}$ from $n=2$ .

Solution. (a) The terms tend to $1/3 \neq 0$ : Divergence Test, diverges.

(b) A rational-looking expression: for large $n$ the term is like $n/(3n^{3}) = 1/(3n^{2})$ , so limit comparison with $1/n^{2}$ gives $L = 1/3$ , and the series converges.

(c) The term is $f(n)$ with $f(x) = x^{2}e^{-x^{3}}$ , positive and decreasing for $x\ge1$ , and integrable by substituting $u = x^{3}$ :

\int_1^{\infty} x^2 e^{-x^3}dx = \frac13\int_1^{\infty}e^{-u}du = \frac{1}{3e}.

Integral Test. Converges. (The Ratio Test also works, giving $L=0$ .)

(d) Alternating with $b_n = 1/\ln n$ decreasing to $0$ : the Alternating Series Test gives convergence. Since $1/\ln n > 1/n$ , the absolute series diverges, so the convergence is conditional.□

Example 9.78 (Two that look alike). Decide the convergence of $\sum_{n=1}^{\infty} \dfrac{1}{2 + 3^n}$ and $\sum_{n=1}^{\infty} \dfrac{1}{2 + 3n}$ .

Solution. The first has an exponential in the denominator: compare with the geometric series $\sum 3^{-n}$ . Since $1/(2+3^n) < 1/3^n$ and $\sum (1/3)^n$ converges, so does the series.

The second has a linear denominator: compare with the harmonic series. Limit comparison against $b_n = 1/n$ gives $n/(2+3n) \to 1/3$ , finite and positive, so the series diverges with $\sum 1/n$ .

One character changes the answer, which is why step 2 of the recipe — identifying the family — comes before any computation.□

Example 9.79 (A series needing two tests). Classify $\sum_{n=1}^{\infty} \dfrac{(-1)^n\, n}{n^{2} + 4}$ .

Solution. First, absolute convergence: $|a_n| = n/(n^{2}+4)$ behaves like $1/n$ , and limit comparison with the harmonic series gives $L = 1$ , so $\sum|a_n|$ diverges. Not absolutely convergent.

Second, the signed series: $b_n = n/(n^{2}+4)$ tends to $0$ , and $f(x) = x/(x^{2}+4)$ has $f'(x) = (4-x^{2})/(x^{2}+4)^2 < 0$ for $x > 2$ , so the terms eventually decrease. The Alternating Series Test gives convergence.

The series converges conditionally.□

Exercises for §9.12 →

9.13Power series: radius and interval of convergence

Everything so far has tested one fixed series. A power series is a whole family at once: a series of powers of $x$ , which converges for some values of $x$ and diverges for others. The set where it converges is the function's domain.

Definition 9.80 (Power series). A power series centred at $a$ is a series of the form

\sum_{n=0}^{\infty} c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots,

where the $c_n$ are constants called the coefficients. When $a = 0$ it is simply $\sum c_n x^n$ . By convention $(x-a)^0 = 1$ even at $x = a$ , so every power series converges at its centre.

Theorem 9.81 (Convergence of a power series). For the power series $\sum c_n(x-a)^n$ exactly one of three things happens:

it converges only at $x = a$ ;
it converges for every real $x$ ;
there is a number $R > 0$ such that the series converges absolutely for $|x - a| < R$ and diverges for $|x-a| > R$ .

The number $R$ is the radius of convergence, with $R = 0$ in case 1 and $R = \infty$ in case 2. The interval of convergence is the set of $x$ where the series converges: one of $(a-R, a+R)$ , $[a-R, a+R)$ , $(a-R, a+R]$ or $[a-R, a+R]$ , according to what happens at each endpoint.

Proof. The heart of the matter is this: if the series converges at some $x_1$ with $x_1 \neq a$ , then it converges absolutely for every $x$ with $|x-a| < |x_1-a|$ . Indeed, convergence at $x_1$ forces $c_n(x_1-a)^n \to 0$ , so those terms are bounded by some $M$ . Then for such an $x$ ,

|c_n(x-a)^n| = |c_n(x_1-a)^n|\left|\frac{x-a}{x_1-a}\right|^n \le M\,t^n, \qquad t = \left|\frac{x-a}{x_1-a}\right| < 1,

and comparison with the convergent geometric series $M\sum t^n$ gives absolute convergence. Contrapositively, divergence at $x_2$ forces divergence for every $x$ farther from $a$ than $x_2$ . So the set of $x$ where the series converges is an interval about $a$ , and taking $R$ to be half its length (the least upper bound of $|x-a|$ over that set) gives the statement. The endpoints are left undecided because the argument needs a strict inequality.∎

The radius is almost always found by applying the Ratio Test to $\sum |c_n(x-a)^n|$ and demanding that the resulting limit be less than $1$ . The endpoints are then substituted one at a time and the two resulting numerical series are tested by hand — the Ratio Test cannot help there, because it returns $L = 1$ at exactly those points.

Method 9.82 (Finding the interval of convergence).

Apply the Ratio Test to $|a_n| = |c_n(x-a)^n|$ and simplify the limit to the form $|x-a|\cdot K$ .
Solve $|x-a|K < 1$ for $|x-a|$ ; the bound is $R$ .
Substitute $x = a - R$ into the original series and test the resulting numerical series.
Substitute $x = a + R$ and test that one.
Report the interval, with a square bracket at each endpoint that converged.

Intuition. A power series is an infinite polynomial, and like a polynomial it is only as good as the numbers you feed it. The radius measures how far from the centre the terms still shrink fast enough for the sum to settle. Inside that distance the powers $(x-a)^n$ beat the coefficients and everything converges; outside it the powers win the other way and the series explodes. At exactly the radius the two effects are in balance, which is why the endpoints have to be examined by hand and why they can behave differently from one another.

Example 9.83 (A radius and both endpoints divergent). Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{2^n}$ .

Solution. Ratio Test on absolute values:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{n+1}}{2^{n+1}}\cdot\frac{2^n}{|x|^n} = \frac{|x|}{2}.

This is less than $1$ exactly when $|x| < 2$ , so $R = 2$ and the centre is $0$ .

At $x = 2$ : the series is $\sum 2^n/2^n = \sum 1$ , whose terms do not tend to $0$ ; it diverges. At $x = -2$ : the series is $\sum(-1)^n$ , which also has terms not tending to $0$ and diverges.

Interval of convergence: $(-2, 2)$ . (Inside it the sum is geometric: $\sum (x/2)^n = \frac{x/2}{1 - x/2}$ from $n=1$ .)□

Example 9.84 (One endpoint in, one out). Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x-1)^n}{n}$ .

Solution. Ratio Test:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-1|^{n+1}}{n+1}\cdot\frac{n}{|x-1|^n} = |x-1|\cdot\frac{n}{n+1} \longrightarrow |x-1|.

So the series converges for $|x-1| < 1$ , giving $R = 1$ and centre $a = 1$ : the candidate interval is $(0,2)$ .

At $x = 2$ : $\sum 1/n$ , the harmonic series, diverges. At $x = 0$ : $\sum (-1)^n/n$ , which converges by the Alternating Series Test ( $1/n$ decreasing to $0$ ).

Interval of convergence: $[0, 2)$ .□

Example 9.85 (Both endpoints convergent). Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (x+2)^n}{n^{2} 3^n}$ .

Solution. Ratio Test:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x+2|^{n+1}}{(n+1)^2 3^{n+1}}\cdot\frac{n^2 3^n}{|x+2|^n} = \frac{|x+2|}{3}\left(\frac{n}{n+1}\right)^2 \longrightarrow \frac{|x+2|}{3}.

Convergence needs $|x+2| < 3$ , so $R = 3$ about the centre $a = -2$ : the candidate interval is $(-5, 1)$ .

At $x = 1$ : the term becomes $(-1)^n 3^n/(n^{2} 3^n) = (-1)^n/n^{2}$ , which converges absolutely ( $p = 2$ ). At $x = -5$ : the term becomes $(-1)^n(-3)^n/(n^{2} 3^n) = 1/n^{2}$ , a convergent $p$ -series.

Interval of convergence: $[-5, 1]$ .□

Example 9.86 (Infinite and zero radius). Find the intervals of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and $\displaystyle\sum_{n=0}^{\infty} n!\,x^n$ .

Solution. For the first,

\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{n+1}}{(n+1)!}\cdot\frac{n!}{|x|^n} = \frac{|x|}{n+1}\longrightarrow 0

for every $x$ . The limit is always below $1$ , so $R = \infty$ and the interval is $(-\infty,\infty)$ . (This series is $e^x$ .)

For the second, with $x \neq 0$ ,

\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!|x|^{n+1}}{n!|x|^n} = (n+1)|x| \longrightarrow \infty,

so it diverges for every $x \neq 0$ and converges only at $x = 0$ : $R = 0$ , interval $\{0\}$ .□

Example 9.87 (When only even powers appear). Find the radius and interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{4^n}$ .

Solution. Consecutive terms differ by $x^{2}$ , not $x$ , so the ratio is

\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{2n+2}}{4^{n+1}}\cdot\frac{4^n}{|x|^{2n}} = \frac{x^2}{4}.

This is less than $1$ when $x^{2} < 4$ , that is $|x| < 2$ , so $R = 2$ . At $x = \pm 2$ the term is $(-1)^n 4^n/4^n = (-1)^n$ , which does not tend to $0$ , so both endpoints diverge.

Interval: $(-2,2)$ . Note that the radius came from solving $x^{2} < 4$ , not from reading the limit directly — a step easy to skip when only even powers appear.□

Pitfall. The endpoints are not optional. The three examples above produced $(-2,2)$ , $[0,2)$ and $[-5,1]$ , all with the same kind of ratio computation; nothing but a direct test at $x = a \pm R$ distinguishes them. And a numerical series at an endpoint needs a real test: "it is alternating" is not enough without $b_n \downarrow 0$ .

Exercises for §9.13 →

9.14Representing functions as power series

A power series defines a function on its interval of convergence, and inside that interval it behaves exactly like a polynomial: it can be differentiated and integrated term by term. That single permission turns the geometric series into a machine for producing series representations of unfamiliar functions.

Theorem 9.88 (Term-by-term differentiation and integration). Suppose $\sum c_n(x-a)^n$ has radius of convergence $R > 0$ and let $f(x)$ be its sum on $(a-R, a+R)$ . Then $f$ is differentiable there, and

f'(x) = \sum_{n=1}^{\infty} n\,c_n (x-a)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n (x-a)^{n+1}}{n+1}.

Both new series have the same radius of convergence $R$ , though their behaviour at the endpoints may differ from the original's.

Proof. The full proof is an exercise in uniform convergence and belongs to real analysis. The radius claim, though, is a short computation: the differentiated series has coefficients $n c_n$ , and since $\sqrt[n]{n} \to 1$ , the root test limit $\limsup \sqrt[n]{n|c_n|}$ equals $\limsup\sqrt[n]{|c_n|}$ . The same numbers control both radii, so both are $R$ .∎

The starting point for everything is the geometric series, read as an identity between a function and a series:

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots, \qquad |x| < 1.

Method 9.89 (Building a series from a known one).

Substitute: replace $x$ by another expression (such as $-x^{2}$ or $x/3$ ) and translate the convergence condition accordingly.
Multiply by a power of $x$ or by a constant.
Differentiate term by term.
Integrate term by term, then fix the constant by evaluating at one point.
Add or multiply two known series.

Intuition. Five series — for $1/(1-x)$ , $e^x$ , $\sin x$ , $\cos x$ and $\ln(1+x)$ — plus these five moves generate essentially every series a calculus course asks for. Computing $n$ derivatives of $\arctan x$ by hand is a miserable afternoon; recognising $\arctan' x = 1/(1+x^{2})$ as a geometric series in $-x^{2}$ and integrating takes two lines.

Example 9.90 (Substitution). Find a power series for $\dfrac{1}{1+4x^{2}}$ and its interval of convergence.

Solution. Write the function as $\dfrac{1}{1 - (-4x^{2})}$ and substitute $-4x^{2}$ for $x$ in the geometric series:

\frac{1}{1+4x^2} = \sum_{n=0}^{\infty} (-4x^2)^n = \sum_{n=0}^{\infty} (-1)^n 4^n x^{2n} = 1 - 4x^2 + 16x^4 - \cdots

The condition $|{-4x^{2}}| < 1$ becomes $x^{2} < 1/4$ , that is $|x| < 1/2$ . At $x = \pm 1/2$ the terms are $(-1)^n$ , which do not tend to $0$ , so the interval is $(-1/2, 1/2)$ .□

Example 9.91 (Differentiating a known series). Find a power series for $\dfrac{1}{(1-x)^2}$ and use it to evaluate $\sum_{n=1}^{\infty} \dfrac{n}{2^n}$ .

Solution. Differentiate the geometric series term by term:

\frac{d}{dx}\frac{1}{1-x} = \frac{1}{(1-x)^2}, \qquad \frac{d}{dx}\sum_{n=0}^{\infty}x^n = \sum_{n=1}^{\infty} n x^{n-1},

so $\dfrac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} = 1 + 2x + 3x^{2} + \cdots$ for $|x|<1$ .

To get the requested sum, multiply by $x$ : $\dfrac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n$ . Setting $x = 1/2$ (legal, since $|1/2| < 1$ ):

\sum_{n=1}^{\infty}\frac{n}{2^n} = \frac{1/2}{(1/2)^2} = 2.

Check: $1/2 + 2/4 + 3/8 + 4/16 + 5/32 \approx 1.78$ , climbing toward $2$ .□

Example 9.92 (Integrating to get the logarithm). Derive the Maclaurin series for $\ln(1+x)$ and give its interval of convergence.

Solution. Start from the geometric series with $-t$ in place of $x$ :

\frac{1}{1+t} = \sum_{n=0}^{\infty}(-1)^n t^n, \qquad |t| < 1.

Integrate both sides from $0$ to $x$ , which is legal for $|x| < 1$ :

\ln(1+x) = \int_0^x \frac{dt}{1+t} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots

The constant of integration is $0$ because both sides vanish at $x = 0$ . Reindexing, $\ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n-1}x^n/n$ . The radius is still $1$ ; at $x = 1$ the series is the convergent alternating harmonic series, and at $x = -1$ it is $-\sum 1/n$ , which diverges. So the interval is $(-1, 1]$ , and setting $x = 1$ gives the striking identity

\ln 2 = 1 - \frac12 + \frac13 - \frac14 + \cdots

□

Example 9.93 (Integrating to get the arctangent). Derive the Maclaurin series for $\arctan x$ .

Solution. Substitute $-t^{2}$ into the geometric series:

\frac{1}{1+t^2} = \sum_{n=0}^{\infty}(-1)^n t^{2n}, \qquad |t| < 1.

Since $\arctan x = \int_0^x dt/(1+t^{2})$ , integrate term by term:

\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}+\cdots

for $|x| < 1$ ; both endpoints converge by the Alternating Series Test, so the interval is $[-1,1]$ . At $x = 1$ this gives Leibniz's formula $\pi/4 = 1 - 1/3 + 1/5 - \cdots$ , beautiful but useless for computation: the error bound $1/(2N+1)$ means that an error below $0.001$ needs $500$ terms.□

Example 9.94 (Using a series to sum a number series). Evaluate $\sum_{n=1}^{\infty} \dfrac{1}{n\,2^n}$ .

Solution. Recognise the shape: $\sum x^n/n$ with $x = 1/2$ . From the logarithm series with $-x$ in place of $x$ ,

-\ln(1-x) = \sum_{n=1}^{\infty}\frac{x^n}{n}, \qquad |x| < 1.

At $x = 1/2$ this gives $-\ln(1/2) = \ln 2$ . So the sum is $\ln 2 \approx 0.693$ . Check: $0.5 + 0.125 + 0.0417 + 0.0156 + 0.00625 \approx 0.688$ , converging on $0.693$ .□

Pitfall. Term-by-term differentiation and integration are guaranteed only inside the radius of convergence. The endpoints can change: $\sum x^n$ has interval $(-1,1)$ , but its integral $\sum x^{n+1}/(n+1)$ converges at $x = -1$ as well. Always retest the endpoints after operating on a series.

Exercises for §9.14 →

9.15Taylor and Maclaurin series

The previous section built series for functions that happened to be disguised geometric series. Taylor's idea is to go the other way: assume a function has a power series representation and read off what the coefficients must be.

Suppose $f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n$ on some interval about $a$ . Setting $x = a$ kills every term but the first, so $c_0 = f(a)$ . Differentiating term by term and setting $x = a$ gives $c_1 = f'(a)$ . Differentiating twice gives $f''(a) = 2c_2$ , and after $n$ differentiations the constant term is $n!\,c_n$ . So the coefficients are forced.

Definition 9.95 (Taylor and Maclaurin series). The Taylor series of $f$ centred at $a$ is

\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots,

defined whenever $f$ has derivatives of all orders at $a$ . The case $a = 0$ is the Maclaurin series. The $n$ th Taylor polynomial $T_n(x)$ is the partial sum through the $(x-a)^n$ term.

Theorem 9.96 (Uniqueness of power series coefficients). If $f(x) = \sum c_n(x-a)^n$ for all $x$ in some interval about $a$ , then $c_n = f^{(n)}(a)/n!$ for every $n$ . A function has at most one power series representation about a given centre.

Proof. Differentiate the series $n$ times term by term (legal inside the radius of convergence by the term-by-term theorem). Every term of degree below $n$ is annihilated, and every term of degree above $n$ retains a factor $(x-a)$ . Setting $x = a$ leaves $f^{(n)}(a) = n!\,c_n$ .∎

Uniqueness is what licenses all the shortcuts of the previous section: a series obtained by substituting into or integrating a known one is the Taylor series, so there is never any need to compute derivatives when a shortcut exists.

Pitfall. A function can have a Taylor series that converges but does not equal the function. The standard example is $f(x) = e^{-1/x^{2}}$ for $x \neq 0$ with $f(0)=0$ : every derivative at $0$ is $0$ , so its Maclaurin series is $0 + 0x + 0x^{2} + \cdots = 0$ , which converges everywhere and agrees with $f$ only at $x = 0$ . Writing $f(x) = \sum f^{(n)}(a)(x-a)^n/n!$ therefore requires proof, and the tool for that proof is the remainder.

Intuition. Building a Taylor polynomial is matching the function at one point with increasing fidelity. The constant term matches the value; the linear term matches the slope; the quadratic term matches the curvature; each further term matches one more derivative. Near the centre those matches accumulate into an excellent copy of the function — which is precisely how a calculator computes $\sin(0.5)$ : not by geometry but by adding $0.5 - 0.5^3/6 + 0.5^5/120$ and stopping when the next term is below the display precision.

The standard Maclaurin series, all derived in this chapter or by the same methods, are worth memorising.

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x|<1; \qquad e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad \text{all } x;

\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \text{all } x; \qquad \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}, \quad \text{all } x;

\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}, \quad -1 < x \le 1; \qquad \arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x|\le 1.

Example 9.97 (A Maclaurin series from the definition). Find the Maclaurin series for $f(x) = e^x$ and its radius of convergence.

Solution. Every derivative of $e^x$ is $e^x$ , so $f^{(n)}(0) = e^{0} = 1$ for every $n$ , and

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

(pending the proof below that the series really equals $e^x$ ). The Ratio Test gives $|x|/(n+1) \to 0$ for every $x$ , so $R = \infty$ . Check at $x=1$ : $1 + 1 + 0.5 + 0.1667 + 0.0417 + 0.0083 = 2.7167$ , approaching $e = 2.71828$ .□

Example 9.98 (A Taylor series about a nonzero centre). Find the Taylor series of $f(x) = \ln x$ centred at $a = 1$ .

Solution. Differentiate repeatedly: $f'(x) = x^{-1}$ , $f''(x) = -x^{-2}$ , $f'''(x) = 2x^{-3}$ , $f^{(4)}(x) = -6x^{-4}$ , and in general $f^{(n)}(x) = (-1)^{n-1}(n-1)!\,x^{-n}$ for $n \ge 1$ . At $x = 1$ this is $(-1)^{n-1}(n-1)!$ , so the coefficient is

\frac{f^{(n)}(1)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n!} = \frac{(-1)^{n-1}}{n}.

With $f(1) = 0$ ,

\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n, \qquad 0 < x \le 2.

This agrees with the series for $\ln(1+u)$ under $u = x-1$ , as uniqueness demands.□

Example 9.99 (Series by substitution and multiplication). Find the Maclaurin series for $x\cos(x^{2})$ through the $x^{9}$ term.

Solution. Substitute $x^{2}$ into the cosine series:

\cos(x^2) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{4n}}{(2n)!} = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots

Multiply by $x$ :

x\cos(x^2) = x - \frac{x^5}{2} + \frac{x^9}{24} - \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{4n+1}}{(2n)!}.

Computing nine derivatives of $x\cos(x^{2})$ would have taken an hour; uniqueness says this answer is the same one.□

Two series can also be multiplied or divided, at least to as many terms as you need. The justification is that inside a common interval of convergence, both are genuine functions, and the product's Taylor series is unique.

Example 9.100 (Multiplying two series). Find the Maclaurin series for $e^x\sin x$ through the $x^{4}$ term.

Solution. Write both factors out to enough terms that every product of total degree at most $4$ is available:

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots, \qquad \sin x = x - \frac{x^3}{6}+\cdots

Collect by degree. Degree $1$ : $1\cdot x = x$ . Degree $2$ : $x\cdot x = x^{2}$ . Degree $3$ : $1\cdot(-x^{3}/6) + (x^{2}/2)\cdot x = -\tfrac16 x^{3} + \tfrac12 x^{3} = \tfrac13 x^{3}$ . Degree $4$ : $x\cdot(-x^{3}/6) + (x^{3}/6)\cdot x = -\tfrac16 x^{4} + \tfrac16 x^{4} = 0$ . So

e^x\sin x = x + x^2 + \frac{x^3}{3} + 0\cdot x^4 + \cdots

Check numerically at $x = 0.2$ : the polynomial gives $0.2 + 0.04 + 0.00267 = 0.24267$ , while $e^{0.2}\sin(0.2) = 1.2214\times 0.19867 = 0.24265$ .□

Example 9.101 (Dividing two series). Find the Maclaurin series for $\tan x$ through the $x^{3}$ term.

Solution. Use $\tan x = \sin x/\cos x$ and long division of series. Write $\tan x = c_1x + c_3x^{3} + \cdots$ (only odd powers, since $\tan$ is odd) and require $\tan x\cdot\cos x = \sin x$ :

(c_1x + c_3x^3+\cdots)\left(1 - \frac{x^2}{2}+\cdots\right) = x - \frac{x^3}{6}+\cdots

Matching the coefficient of $x$ : $c_1 = 1$ . Matching the coefficient of $x^{3}$ : $c_3 - c_1/2 = -1/6$ , so $c_3 = 1/2 - 1/6 = 1/3$ . Hence

\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15}+\cdots

Check at $x = 0.3$ : $0.3 + 0.009 = 0.309$ versus $\tan(0.3) = 0.30934$ .□

The last standard family generalises the binomial theorem from whole-number exponents to arbitrary ones.

Theorem 9.102 (Binomial series). For any real number $k$ and $|x| < 1$ ,

(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n, \qquad \binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!},

with $\binom{k}{0} = 1$ . If $k$ is a nonnegative integer the coefficients vanish beyond $n = k$ and the series is the ordinary binomial theorem.

Proof. Let $f(x) = (1+x)^k$ . Then $f^{(n)}(x) = k(k-1)\cdots(k-n+1)(1+x)^{k-n}$ , so $f^{(n)}(0) = k(k-1)\cdots(k-n+1)$ and the Maclaurin coefficients are exactly $\binom{k}{n}$ . The Ratio Test gives

\left|\frac{c_{n+1}x^{n+1}}{c_nx^n}\right| = \left|\frac{k-n}{n+1}\right||x| \longrightarrow |x|,

so the radius is $1$ . That the series actually sums to $(1+x)^k$ follows by checking that its sum $g$ satisfies the differential equation $(1+x)g'(x) = k\,g(x)$ with $g(0)=1$ , which has the unique solution $(1+x)^k$ .∎

Example 9.103 (Square root as a binomial series). Find the Maclaurin series for $\sqrt{1+x}$ through the $x^{3}$ term.

Solution. Take $k = 1/2$ . The coefficients are

\binom{1/2}{1} = \frac12, \qquad \binom{1/2}{2} = \frac{\frac12\left(-\frac12\right)}{2} = -\frac18, \qquad \binom{1/2}{3} = \frac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{6} = \frac{1}{16}.

\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots, \qquad |x|<1.

Check at $x = 0.2$ : $1 + 0.1 - 0.005 + 0.0005 = 1.0955$ against $\sqrt{1.2} = 1.09545$ .□

Example 9.104 (Forcing a binomial series into shape). Find the Maclaurin series for $\dfrac{1}{\sqrt{4-x}}$ and its radius of convergence.

Solution. Factor out the constant so the bracket reads $1 + (\text{something})$ :

\frac{1}{\sqrt{4-x}} = \frac{1}{\sqrt{4\left(1 - \frac{x}{4}\right)}} = \frac12\left(1 - \frac{x}{4}\right)^{-1/2}.

Apply the binomial series with $k = -1/2$ and $-x/4$ in place of $x$ . Since

\binom{-1/2}{n} = \frac{(-\frac12)(-\frac32)\cdots(-\frac{2n-1}{2})}{n!} = \frac{(-1)^n (2n)!}{4^n (n!)^2},

we get

\frac{1}{\sqrt{4-x}} = \frac12\sum_{n=0}^{\infty}\frac{(2n)!}{16^n (n!)^2}x^n = \frac12 + \frac{x}{16} + \frac{3x^2}{256}+\cdots

Convergence requires $|x/4| < 1$ , so $R = 4$ . Check at $x=1$ : $0.5 + 0.0625 + 0.0117 = 0.5742$ against $1/\sqrt 3 = 0.57735$ ; the next term, $5/2048 \approx 0.0024$ , closes most of the gap.□

Exercises for §9.15 →

9.16Taylor's Inequality and error bounds

A Taylor polynomial is only useful with a guarantee attached. The guarantee comes from the remainder.

Definition 9.105 (Taylor remainder). For a function $f$ with $n+1$ derivatives near $a$ , the $n$ th remainder is

R_n(x) = f(x) - T_n(x),

the exact error made by replacing $f$ with its $n$ th Taylor polynomial. The Taylor series of $f$ converges to $f(x)$ precisely when $R_n(x) \to 0$ as $n \to \infty$ .

Theorem 9.106 (Taylor's Theorem, Lagrange form). If $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$ , then there is a number $z$ strictly between $a$ and $x$ with

R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}.

Proof. This is the Mean Value Theorem applied $n+1$ times. For $n = 0$ it reads $f(x) - f(a) = f'(z)(x-a)$ , which is the Mean Value Theorem itself. In general, fix $x$ and define $g(t) = f(x) - T_n(t;x)$ where $T_n(t;x)$ is the Taylor polynomial of $f$ about $t$ evaluated at $x$ ; repeated application of the generalised Mean Value Theorem to $g$ against $(x-t)^{n+1}$ produces the stated formula. The detailed bookkeeping belongs to a first analysis course.∎

Theorem 9.107 (Taylor's Inequality). If $|f^{(n+1)}(t)| \le M$ for all $t$ between $a$ and $x$ , then

|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}.

Proof. Immediate from the Lagrange form: the unknown $z$ lies between $a$ and $x$ , so $|f^{(n+1)}(z)| \le M$ .∎

The bound has two competing pieces. The factor $|x-a|^{n+1}$ says accuracy degrades as you move away from the centre, which is why Taylor polynomials are local tools. The factor $(n+1)!$ in the denominator says accuracy improves explosively with degree, because factorials outrun every power. For $|x-a| \le 1$ the $10$ th-degree bound already carries $1/11! \approx 2.5\times10^{-8}$ .

Intuition. Taylor's Inequality is an insurance policy on an approximation. You do not know the true error, but you know the worst the $(n+1)$ st derivative can be on the stretch between the centre and your point, and that worst case buys a guarantee. The guarantee is usually pessimistic — the real error is often an order of magnitude smaller — which is fine: a bound you can compute beats an error you cannot.

Theorem 9.108 (The exponential equals its series). For every real $x$ , $\displaystyle e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ .

Proof. Fix $x$ and let $d = |x|$ . Every derivative of $e^t$ is $e^t$ , which on the interval between $0$ and $x$ is bounded by $M = e^{d}$ . Taylor's Inequality gives

|R_n(x)| \le \frac{e^{d}}{(n+1)!}d^{\,n+1}.

Now $d^{n+1}/(n+1)! \to 0$ as $n\to\infty$ for every fixed $d$ , because it is the general term of the convergent series $\sum d^n/n!$ (Ratio Test limit $0$ ), and terms of a convergent series tend to $0$ . So $R_n(x)\to0$ and the series converges to $e^x$ .∎

Corollary 9.109 (The sine equals its series). For every real $x$ , $\displaystyle \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ , and likewise for $\cos x$ .

Proof. Every derivative of $\sin$ is $\pm\sin$ or $\pm\cos$ , so $|f^{(n+1)}(t)| \le 1 = M$ everywhere. Taylor's Inequality gives $|R_n(x)| \le |x|^{n+1}/(n+1)! \to 0$ for every $x$ , exactly as above. The Maclaurin coefficients of $\sin$ cycle $0, 1, 0, -1$ , producing the stated odd-power series.∎

Example 9.110 (Bounding the error of a specific approximation). Estimate $e$ by $T_3(1)$ for $f(x)=e^x$ about $0$ , and bound the error using $e < 3$ .

Solution. The polynomial is $T_3(x) = 1 + x + x^{2}/2 + x^{3}/6$ , so $T_3(1) = 1+1+0.5+0.1\overline{6} = 8/3 \approx 2.667$ . For the bound, $f^{(4)}(t) = e^t \le e^{1} < 3$ on $[0,1]$ , so with $M = 3$ , $n = 3$ and $|x - a| = 1$ :

|R_3(1)| \le \frac{3}{4!}\cdot1^4 = \frac{3}{24} = \frac18 = 0.125.

The true error is $2.71828 - 2.66667 = 0.0516$ , comfortably inside the guarantee — a reminder that the bound is an upper limit, not an estimate.□

Example 9.111 (Choosing the degree to meet a tolerance). How many terms of the Maclaurin series for $e^x$ guarantee $e^{0.5}$ to within $0.0001$ ?

Solution. On $[0, 0.5]$ , $|f^{(n+1)}(t)| = e^t \le e^{0.5} < 2$ , so take $M = 2$ :

|R_n(0.5)| \le \frac{2\,(0.5)^{n+1}}{(n+1)!}.

Test values of $n$ : $n=4$ gives $2(0.03125)/120 = 5.2\times10^{-4}$ , too big; $n=5$ gives $2(0.015625)/720 = 4.3\times10^{-5} < 10^{-4}$ . So $n = 5$ works: six terms, through $x^{5}$ . Their sum is $1.6487$ , and $e^{0.5} = 1.64872$ .□

Example 9.112 (Sine near its centre). Show that $T_3(x) = x - x^{3}/6$ approximates $\sin x$ to within $0.0001$ when $|x| \le 0.3$ , and compare with the alternating-series bound.

Solution. Since $\sin$ has $T_4 = T_3$ (the $x^{4}$ coefficient is $0$ ), use $n = 4$ , which gives a sharper bound. With $M = 1$ :

|R_4(x)| \le \frac{|x|^5}{5!} \le \frac{0.3^5}{120} = \frac{0.00243}{120} = 2.03\times10^{-5} < 0.0001.

The alternating series estimation gives the same number: the first omitted term of $x - x^{3}/6 + x^{5}/120 - \cdots$ is $x^{5}/120$ . For alternating Taylor series the two methods agree, and the alternating one is less work.□

Pitfall. $M$ must bound the $(n+1)$ st derivative on the whole interval between the centre and the point, not just at one end. For $f(x)=e^x$ on $[0,2]$ the bound is $e^{2} \approx 7.4$ , not $1$ . Choosing $M$ too small produces a guarantee that is simply false.

Exercises for §9.16 →

9.17Applications of Taylor polynomials

Taylor series earn their place in three ways: they compute integrals that have no elementary antiderivative, they settle limits without repeated use of L'Hôpital's Rule, and they justify the simplified formulas that physics uses in the small-quantity regime.

Example 9.113 (An integral with no elementary antiderivative). Approximate $\displaystyle\int_0^1 e^{-x^{2}}\,dx$ to three decimal places.

Solution. No elementary function differentiates to $e^{-x^{2}}$ , so integrate the series instead. Substituting $-x^{2}$ into the exponential series,

e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \frac{x^8}{24} - \cdots,

valid for all $x$ . Integrating term by term from $0$ to $1$ :

\int_0^1 e^{-x^2}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)} = 1 - \frac13 + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \frac{1}{1320} + \frac{1}{9360}-\cdots

This alternates with decreasing terms, so the error after a partial sum is at most the next term. To get three decimals we need a term below $0.0005$ : $1/1320 \approx 0.00076$ is not small enough, $1/9360 \approx 0.000107$ is. Summing through $1/9360$ :

1 - 0.33333 + 0.1 - 0.02381 + 0.00463 - 0.00076 + 0.00011 = 0.74684.

So the integral is $0.747$ to three decimal places. (The true value is $0.746824\ldots$ )□

Example 9.114 (A limit by series). Compute $\displaystyle\lim_{x\to0}\frac{e^x - 1 - x}{x^{2}}$ .

Solution. Substitute the series for $e^x$ into the numerator:

e^x - 1 - x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}+\cdots\right) - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots

Divide by $x^{2}$ :

\frac{e^x-1-x}{x^2} = \frac12 + \frac{x}{6} + \frac{x^2}{24}+\cdots \longrightarrow \frac12

as $x \to 0$ , since a power series is continuous inside its interval of convergence. Two applications of L'Hôpital's Rule give the same answer, but the series also shows how the quotient approaches $1/2$ : linearly in $x$ , with slope $1/6$ .□

Example 9.115 (A harder limit). Compute $\displaystyle\lim_{x\to0}\frac{\sin x - x + x^{3}/6}{x^{5}}$ .

Solution. The numerator is $\sin x$ minus its own third-degree Taylor polynomial, so it is exactly the tail of the sine series:

\sin x - x + \frac{x^3}{6} = \frac{x^5}{120} - \frac{x^7}{5040} + \cdots

Dividing by $x^{5}$ gives $\dfrac{1}{120} - \dfrac{x^{2}}{5040}+\cdots \to \dfrac{1}{120}$ . Doing this by L'Hôpital's Rule would take five differentiations of numerator and denominator.□

Example 9.116 (Relativistic kinetic energy). In special relativity the kinetic energy of a mass $m$ at speed $v$ is

K = mc^2\left(\frac{1}{\sqrt{1 - v^2/c^2}} - 1\right).

Show that for $v$ much smaller than $c$ this reduces to the Newtonian $\tfrac12mv^{2}$ , and estimate the error at $v = 100$ m/s.

Solution. Let $u = v^{2}/c^{2}$ , a small positive number, and apply the binomial series with $k = -1/2$ :

(1-u)^{-1/2} = 1 + \frac{u}{2} + \frac{3u^2}{8} + \frac{5u^3}{16}+\cdots

Therefore

K = mc^2\left(\frac{u}{2} + \frac{3u^2}{8}+\cdots\right) = \frac12 m c^2\frac{v^2}{c^2} + \frac38 mc^2\frac{v^4}{c^4}+\cdots = \frac12 mv^2 + \frac38\frac{mv^4}{c^2}+\cdots

The leading term is the Newtonian kinetic energy; relativity corrects it by a term smaller by a factor of order $v^{2}/c^{2}$ . At $v = 100$ m/s with $c = 3\times10^8$ m/s, $v^{2}/c^{2} \approx 1.1\times10^{-13}$ , so the correction is about $10^{-13}$ of the Newtonian value: entirely invisible, which is why Newton's formula stood for two centuries.□

Example 9.117 (Small-angle approximation in optics). The path difference in a lens calculation involves $\cos\theta$ for small $\theta$ . Replace $\cos\theta$ by $1 - \theta^{2}/2$ and bound the error for $|\theta| \le 0.1$ radians.

Solution. The Maclaurin series is $\cos\theta = 1 - \theta^{2}/2 + \theta^{4}/24 - \cdots$ , so the proposed approximation is $T_3$ (and also $T_2$ ). By the alternating series estimate the error is at most the first omitted term,

\frac{\theta^4}{24} \le \frac{(0.1)^4}{24} = \frac{10^{-4}}{24} \approx 4.2\times10^{-6}.

So for angles up to about $6$ degrees the quadratic replacement is accurate to five decimal places — the justification for every "small-angle approximation" in physics. Dropping one term further, to $\cos\theta \approx 1$ , costs $\theta^{2}/2 = 0.005$ , a thousand times worse.□

Intuition. Each of these applications follows the same move: replace a function you cannot handle by a polynomial you can, then carry an error bound along so the answer means something. The bound is what separates an approximation from a guess. A physicist writing $\sin\theta \approx \theta$ is silently invoking $|R| \le \theta^{3}/6$ , and would notice immediately if $\theta$ grew large enough to make that number matter.

Pitfall. Series manipulations are valid only where the series converge. Integrating the series for $1/(1-x)$ from $0$ to $2$ is meaningless, since the representation fails past $x = 1$ . Always check that the interval of integration or the point of evaluation lies inside the interval of convergence.

Summary. $\sum a_n = S$ means the partial sums $s_N = \sum_{n=1}^N a_n$ converge to $S$ , and every bounded monotone sequence converges — the fact behind most of the tests. Convergence forces $a_n \to 0$ , but not conversely, so the Divergence Test only proves divergence. Benchmarks: $\sum_{n=0}^\infty ar^n$ converges iff $|r|<1$ , to $\frac{a}{1-r}$ ; $\sum 1/n^p$ converges iff $p>1$ . The tests and their hypotheses: Integral Test for $a_n = f(n)$ with $f$ continuous, positive and decreasing on $[N,\infty)$ , with $\int_{N+1}^\infty f \le R_N \le \int_N^\infty f$ ; Direct Comparison for $0 \le a_n \le b_n$ ; Limit Comparison for $a_n,b_n>0$ with $a_n/b_n \to L$ , $0<L<\infty$ ; the Alternating Series Test when $b_n$ decreases to $0$ , giving $|S-s_N| \le b_{N+1}$ ; Ratio and Root, absolutely convergent for $L<1$ , divergent for $L>1$ , silent at $L=1$ . Absolute convergence implies convergence. A power series $\sum c_n(x-a)^n$ converges at $a$ only, everywhere, or absolutely on $|x-a|<R$ with the endpoints tested by hand; inside $R$ it may be differentiated and integrated term by term with the same $R$ . Coefficients are forced: $c_n = f^{(n)}(a)/n!$ . The Taylor series equals $f$ exactly when $R_n(x)\to 0$ , which Taylor's Inequality $|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$ , valid when $|f^{(n+1)}| \le M$ between $a$ and $x$ , delivers for $e^x$ , $\sin x$ and $\cos x$ on all of $\mathbb{R}$ .

Exercises for §9.17 →

Common Mistakes

Concluding that $\sum a_n$ converges because $a_n \to 0$ . The harmonic series is the standing counterexample; the Divergence Test only ever proves divergence.
Using the geometric formula $a/(1-r)$ without checking $|r| < 1$ , or reading off $a$ as the coefficient rather than as the first term actually present in the series.
Applying the Ratio or Root Test to a rational function of $n$ . Both give $L = 1$ there and decide nothing; use a comparison or the Integral Test.
Reporting "diverges" when a test is inconclusive. $L = 1$ in the Ratio Test is silence, not a verdict.
Comparing in the wrong direction: bounding a series above by a divergent series, or below by a convergent one, proves nothing.
Using a comparison or the Integral Test on a series with negative terms. Pass to $\sum|a_n|$ first, or use the Alternating Series Test.
Forgetting to check that $b_n$ is decreasing in the Alternating Series Test, and then quoting the first-omitted-term error bound, which requires the same hypothesis.
Treating the value of $\int_N^\infty f$ as the sum of the series. The Integral Test compares them; it does not equate them.
Skipping the endpoints of a power series. The radius is only half the answer, and the two endpoints can behave differently from each other.
Cancelling "everything" in a telescoping series without writing out $s_N$ ; a gap of two leaves two survivors at each end, not one.
Assuming a function equals its Taylor series without checking $R_n \to 0$ (or citing a standard series). The function $e^{-1/x^{2}}$ shows the two can differ.
Choosing $M$ in Taylor's Inequality from the value of the derivative at one endpoint instead of its maximum over the whole interval.
Dropping the $n!$ in a Taylor coefficient, or centring at $0$ a series that was asked for about $a$ .

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

← Parametric Equations and Polar Coordinates Vectors, Space Curves, and Motion →

Helpful?