The Alternating Series Test

The harmonic series $1 + \tfrac{1}{2} + \tfrac{1}{3} + \dots$ diverges. But flip every second sign, $1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \dots$, and the result converges. When terms alternate in sign, each one partly cancels the one before, and that’s often enough. The alternating series test tells you exactly when this cancelling guarantees convergence.

Key ideas

Alternating series

An alternating series has terms that switch sign every time. It can be written

$$\sum_{n=1}^{\infty} (-1)^{n+1} b_n = b_1 - b_2 + b_3 - b_4 + \dots \qquad\text{or}\qquad \sum_{n=1}^{\infty} (-1)^n b_n = -b_1 + b_2 - b_3 + \dots$$

where every $b_n \gt 0$. Here $b_n$ is the size of the term, without its sign. You’ll also see $\cos(n\pi)$ used as the sign, since $\cos(n\pi) = (-1)^n$ (in radians).

The test

If

1. the sizes $b_n$ are decreasing: $b_{n+1} \le b_n$ (at least from some point on), and
2. $\displaystyle\lim_{n \to \infty} b_n = 0$,

then the alternating series converges.

Why it works

Picture the partial sums for $1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \dots$. You step forward $1$, back $\tfrac{1}{2}$, forward $\tfrac{1}{3}$, back $\tfrac{1}{4}$, and so on. Each step is shorter than the last (condition 1), so the partial sums zig-zag in a narrowing band. The steps shrink to nothing (condition 2), so the band closes in on a single number: the sum.

The partial sums of $1 - \frac{1}{2} + \frac{1}{3} - \dots$ land alternately above and below the sum $\ln 2 \approx 0.693$.

The picture also shows that the sum always lies between any two consecutive partial sums. That fact leads to the alternating series error bound.

The alternating harmonic series

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$$

converges (to $\ln 2$, as you’ll be able to show with Taylor series), even though the harmonic series without the signs diverges.

What the test can’t do

The alternating series test only ever proves convergence. If $b_n$ does not approach $0$, the series diverges, but the reason is the nth term test, and you should say so. If $b_n \to 0$ but $b_n$ isn’t decreasing, the test simply doesn’t apply. The series might converge or diverge.

On the AP exam, write both conditions explicitly: “The terms alternate in sign, $\frac{1}{n}$ is decreasing, and $\displaystyle\lim_{n \to \infty} \frac{1}{n} = 0$, so the series converges by the alternating series test.”

Worked examples

Example 1: The alternating harmonic series

Show that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges, and find its first five partial sums.

Solution. The series alternates with $b_n = \dfrac{1}{n}$.

1. $\dfrac{1}{n+1} \lt \dfrac{1}{n}$, so $b_n$ is decreasing.
2. $\displaystyle\lim_{n \to \infty} \frac{1}{n} = 0$.

So the series converges by the alternating series test. The partial sums are

$$S_1 = 1, \quad S_2 = 0.5, \quad S_3 \approx 0.833, \quad S_4 \approx 0.583, \quad S_5 \approx 0.783$$

They bounce above and below the sum, $\ln 2 \approx 0.693$.

Example 2: Checking “decreasing” with a derivative

Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1}$ converge or diverge?

Solution. The series alternates with $b_n = \dfrac{n}{n^2 + 1}$.

Is $b_n$ decreasing? It isn’t obvious, because both the top and bottom grow. Let $f(x) = \dfrac{x}{x^2 + 1}$:

$$f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \le 0 \quad\text{for } x \ge 1$$

So $b_n$ is decreasing. The limit is $\displaystyle\lim_{n \to \infty} \frac{n}{n^2 + 1} = 0$ (the bottom has the higher power). Both conditions hold, so the series converges by the alternating series test.

Example 3: When the terms don’t shrink to zero

Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n\, 2n}{3n + 1}$ converge or diverge?

Solution. The sizes are $b_n = \dfrac{2n}{3n + 1}$, and

$$\lim_{n \to \infty} \frac{2n}{3n + 1} = \frac{2}{3} \ne 0 .$$

So the terms $a_n = \dfrac{(-1)^n\, 2n}{3n + 1}$ keep jumping between values near $\tfrac{2}{3}$ and $-\tfrac{2}{3}$, and don’t approach $0$. The series diverges by the $n$th term test. (The alternating series test can’t prove divergence, so don’t cite it here.)

Example 4: Decreasing only eventually

Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln n}{n}$ converge or diverge?

Solution. Here $b_n = \dfrac{\ln n}{n}$ (with $b_1 = 0$, which doesn’t matter). Let $f(x) = \dfrac{\ln x}{x}$:

$$f'(x) = \frac{1 - \ln x}{x^2} \lt 0 \quad\text{when } x \gt e$$

So $b_n$ is decreasing for $n \ge 3$. That’s enough, since the first two terms don’t affect convergence. By L’Hôpital’s rule,

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

Both conditions hold (from $n = 3$ on), so the series converges by the alternating series test.

Common mistakes

Saying “the alternating series test fails, so the series diverges.” The test only proves convergence. If $b_n \not\to 0$, cite the $n$th term test for divergence. If $b_n$ isn’t decreasing, you need a different argument.

Checking only that b_n goes to 0. Both conditions are required. Many AP answers lose a point for never mentioning that $b_n$ is decreasing. When it isn’t obvious, use a derivative or compare $b_{n+1}$ with $b_n$.

Including the sign in b_n. $b_n$ is the positive size of the term. For $\frac{(-1)^n n}{n^2 + 1}$, $b_n = \frac{n}{n^2 + 1}$, not $\frac{(-1)^n n}{n^2 + 1}$. The sequence $(-1)^n b_n$ isn’t decreasing, but that’s irrelevant.

Using the test on a series that doesn’t strictly alternate. $\sum \frac{\sin n}{n^2}$ has mixed signs, but not $+,-,+,-$ in order, so this test doesn’t apply. (It converges by absolute convergence.)

Thinking the series “converges absolutely” because of this test. The alternating series test says nothing about $\sum b_n$. The alternating harmonic series converges, but $\sum \frac{1}{n}$ diverges.

Practice

1. (Warm-up) Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ converge or diverge?

Solution

It alternates with $b_n = \dfrac{1}{\sqrt{n}}$, which is decreasing (the square root increases) and has limit $0$. It converges by the alternating series test.

2. (Warm-up) Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n}{n + 1}$ converge or diverge?

Solution

$b_n = \dfrac{n}{n + 1} \to 1 \ne 0$, so the terms don’t approach $0$. The series diverges by the $n$th term test.

3. (Warm-up) Does $\displaystyle\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n^2}$ converge or diverge?

Solution

In radians, $\cos(n\pi) = (-1)^n$, so the series is $\sum \frac{(-1)^n}{n^2}$. It alternates with $b_n = \frac{1}{n^2}$, which is decreasing with limit $0$. It converges by the alternating series test.

4. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{n^2 + 4}$ converge or diverge? Compute $b_1$, $b_2$, and $b_3$ first.

Solution

$b_1 = \tfrac{1}{5} = 0.2$, $b_2 = \tfrac{2}{8} = 0.25$, $b_3 = \tfrac{3}{13} \approx 0.231$. So $b_n$ goes up at first, then down. Check with $f(x) = \dfrac{x}{x^2 + 4}$:

$$f'(x) = \frac{(x^2 + 4) - 2x^2}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2} \lt 0 \quad\text{for } x \gt 2$$

So $b_n$ is decreasing for $n \ge 2$, which is enough. Also $\displaystyle\lim_{n \to \infty} \frac{n}{n^2 + 4} = 0$. The series converges by the alternating series test.

5. (Core) Does $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ converge or diverge?

Solution

It alternates with $b_n = \dfrac{1}{\ln n}$. Since $\ln n$ is increasing and positive for $n \ge 2$, $b_n$ is decreasing, and $\ln n \to \infty$ means $b_n \to 0$. It converges by the alternating series test.

6. (Core) Find $S_5$ for the alternating harmonic series, to three decimal places. Between which two partial sums must the sum lie?

Solution

$$S_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{47}{60} \approx 0.783$$

The sum lies between consecutive partial sums, so it’s between $S_4 = \tfrac{7}{12} \approx 0.583$ and $S_5 \approx 0.783$. (Indeed, $\ln 2 \approx 0.693$.)

7. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{2^n}$ converge or diverge?

Solution

$b_n = \dfrac{n^2}{2^n}$: $b_1 = \tfrac{1}{2}$, $b_2 = 1$, $b_3 = \tfrac{9}{8}$, $b_4 = 1$, $b_5 = \tfrac{25}{32}$, … It rises at first. Compare consecutive terms:

$$\frac{b_{n+1}}{b_n} = \frac{(n+1)^2}{2n^2}$$

This is less than $1$ when $(n+1)^2 \lt 2n^2$, which is true for $n \ge 3$ (at $n = 3$: $16 \lt 18$). So $b_n$ is decreasing from $n = 3$ on. By L’Hôpital’s rule twice, $\displaystyle\lim_{x \to \infty} \frac{x^2}{2^x} = \lim_{x \to \infty} \frac{2}{(\ln 2)^2 2^x} = 0$. The series converges by the alternating series test.

8. (Challenge) Does $\displaystyle\sum_{n=1}^{\infty} (-1)^n \sin\!\left(\frac{1}{n}\right)$ converge or diverge?

Solution

$b_n = \sin\!\left(\frac{1}{n}\right)$. For $n \ge 1$, $\frac{1}{n}$ is in $(0, 1]$, where sine is positive and increasing. As $n$ increases, $\frac{1}{n}$ decreases, so $\sin\!\left(\frac{1}{n}\right)$ decreases. Also $\sin\!\left(\frac{1}{n}\right) \to \sin 0 = 0$. Both conditions hold, so the series converges by the alternating series test.

(Compare with the comparison tests page, where $\sum \sin\!\left(\frac{1}{n}\right)$ without the signs diverges.)

9. (Challenge) Consider the alternating series

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

where the positive terms are $\frac{1}{n}$ for odd $n$ and the negative terms are $\frac{1}{n^2}$ for even $n$. The sizes go to $0$. Explain why the alternating series test doesn’t apply, and show that the series diverges.

Solution

The sizes are not decreasing: for example, $b_2 = \tfrac{1}{4}$ but $b_3 = \tfrac{1}{3} \gt \tfrac{1}{4}$. So condition 1 fails and the test doesn’t apply.

To see that it diverges, look at the partial sums after an even number of terms:

$$S_{2k} = \left(1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2k - 1}\right) - \left(\frac{1}{2^2} + \frac{1}{4^2} + \dots + \frac{1}{(2k)^2}\right)$$

The subtracted part is less than $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$, so it stays bounded. In the first part, each term $\frac{1}{2j - 1}$ is bigger than $\frac{1}{2j}$, so

$$1 + \frac{1}{3} + \dots + \frac{1}{2k - 1} \gt \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2k} = \frac{1}{2}\left(1 + \frac{1}{2} + \dots + \frac{1}{k}\right)$$

That’s half a harmonic partial sum, which grows without bound. So $S_{2k} \to \infty$ and the series diverges. The “decreasing” condition really matters.