Math Core

So far every test has required positive terms. But many important series flip sign with every term, and the cancellation between positive and negative terms can make a series converge even when the positive version would diverge. Alternating series also come with a remarkably simple way to measure how accurate a partial sum is.

Alternating series

An alternating series has terms that switch sign every time. It is usually written as

\sum_{n=1}^{\infty} (-1)^{n+1} a_n = a_1 - a_2 + a_3 - a_4 + \cdots \qquad (a_n > 0),

or with $(-1)^n$ if the first term is negative. The most famous example is the alternating harmonic series,

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

Why the partial sums settle down

Follow the partial sums of the alternating harmonic series: $S_1 = 1$ , then step down by $\frac12$ to $S_2 = 0.5$ , up by $\frac13$ to $S_3 \approx 0.833$ , down by $\frac14$ to $S_4 \approx 0.583$ , and so on. Each step is in the opposite direction from the last and smaller than the last. So the partial sums zigzag back and forth, and the zigzags shrink. If the step sizes shrink to zero, the partial sums squeeze in on a single number. (For this series that number is $\ln 2 \approx 0.693$ .)

Partial sums of 1 − 1/2 + 1/3 − ⋯ zigzag above and below the sum ln 2 ≈ 0.693, with the zigzags shrinking.Open in grapher →

Alternating series test

The alternating series $\displaystyle\sum (-1)^{n+1} a_n$ (with $a_n > 0$ ) converges if both

the terms decrease in size: $a_{n+1} \le a_n$ for all $n$ (at least eventually), and
$\displaystyle\lim_{n\to\infty} a_n = 0$ .

If condition 2 fails, the series diverges by the nth term test. If condition 1 fails, the alternating series test simply doesn't apply, and you need another method.

The alternating series error bound

Since the partial sums bounce back and forth across the true sum $S$ , the sum always lies between two consecutive partial sums. So the error after $n$ terms is less than the size of the next step.

Alternating series error bound

If a series satisfies the conditions of the alternating series test, and $S_n$ is the sum of its first $n$ terms, then

|S - S_n| \le a_{n+1},

the absolute value of the first omitted term. Moreover, the error has the same sign as that first omitted term.

That last sentence tells you whether $S_n$ is an overestimate or an underestimate. If the first omitted term is negative, the true sum is less than $S_n$ , so $S_n$ is an overestimate. If the first omitted term is positive, $S_n$ is an underestimate.

Common mistake

On a free-response question, you must verify both conditions before claiming convergence or using the error bound: the terms decrease in absolute value, and they approach $0$ . Also, the error bound uses the next term $a_{n+1}$ , not the last term you added.

Worked examples

Worked example: Checking the conditions

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ converges.

Solution. Here $a_n = \dfrac{1}{\sqrt n}$ .

$\sqrt{n+1} > \sqrt n$ , so $\dfrac{1}{\sqrt{n+1}} < \dfrac{1}{\sqrt n}$ : the terms decrease.
$\displaystyle\lim_{n\to\infty} \frac{1}{\sqrt n} = 0$ .

By the alternating series test, the series converges, even though $\sum \frac{1}{\sqrt n}$ diverges.

Worked example: When the terms don't shrink to zero

Determine whether $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{2n}{n + 1}$ converges.

Solution. The sizes $\dfrac{2n}{n+1} \to 2$ , not $0$ . So the terms $(-1)^n \dfrac{2n}{n+1}$ do not approach $0$ , and the series diverges by the nth term test.

Worked example: Bounding the error

Let $S = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ . Approximate $S$ with $S_4$ , and bound the error. Is $S_4$ too big or too small?

Solution. The terms $\frac{1}{n^2}$ decrease to $0$ , so the conditions hold.

S_4 = 1 - \frac14 + \frac19 - \frac{1}{16} = \frac{144 - 36 + 16 - 9}{144} = \frac{115}{144} \approx 0.7986.

The first omitted term is $+\dfrac{1}{25}$ , so $|S - S_4| \le \dfrac{1}{25} = 0.04$ . Since the omitted term is positive, $S$ is bigger than $S_4$ : $S_4$ is an underestimate. (In fact $S = \frac{\pi^2}{12} \approx 0.8225$ .)

Worked example: How many terms are needed?

How many terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}$ guarantee an error less than $0.001$ ?

Solution. The terms $\frac{1}{n!}$ decrease to $0$ . We need the first omitted term $\dfrac{1}{(n+1)!}$ to be less than $0.001$ , that is, $(n+1)! > 1000$ . Since $6! = 720$ and $7! = 5040$ , we need $n + 1 = 7$ , so $n = 6$ terms.

Tip

A quick way to check that terms decrease: if $a_n = f(n)$ , show $f'(x) < 0$ . For simple expressions like $\frac{1}{n^2}$ or $\frac{1}{\ln n}$ , "the denominator increases" is enough justification.

Practice

Practice 1

Which of the following best describes $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ ?

Practice 2

Let $S = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ . Using the alternating series error bound, what is the best upper bound on $|S - S_4|$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

What is the smallest number of terms $N$ of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}$ for which the alternating series error bound guarantees $|S - S_N| < 0.001$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

What happens with $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{n + 1}$ ?

Practice 5

Find $S_3$ , the sum of the first three terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

The series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}$ has sum $S$ . For $S_3 = \dfrac23$ from the previous problem, which statement is true?

Practice 7

Find the exact sum of the alternating series $\dfrac12 - \dfrac14 + \dfrac18 - \dfrac1{16} + \cdots$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.