Alternating Series Test

The alternating series test says that an alternating series \sum (-1)^{n+1} b_n, with b_n \gt 0, converges if two conditions hold: the sizes b_n decrease (at least from some point on), and \lim_{n \to \infty} b_n = 0. For example, the alternating harmonic series 1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \cdots converges.

Before this lesson: Infinite Series and Partial Sums, Divergence Test and Convergence of Sequences

Why shrinking back-and-forth steps settle down

An alternating series has terms whose signs switch: +, −, +, −, …. Picture walking along a number line: 1 step forward, 1/2 step back, 1/3 step forward, 1/4 step back, and so on.

Each step undoes part of the one before, but never all of it, because the steps keep getting smaller. So your position zigzags: it overshoots, comes back a little, overshoots by less, and so on. The zigzag gets narrower and narrower, so your position closes in on one point.

If the steps did not shrink to 0, the zigzag would never close up. That is why both conditions matter.

The alternating series test

Suppose b_n \gt 0 for all n. The series \sum (-1)^{n+1} b_n = b_1 - b_2 + b_3 - \cdots (or the same series starting with a minus sign, \sum (-1)^n b_n) converges if:

\text{(1) } b_{n+1} \le b_n \text{ for all } n \text{ (from some point on)} \qquad \text{(2) } \lim_{n \to \infty} b_n = 0

Why it works. The even partial sums S_2, S_4, S_6, \ldots go up, the odd ones S_1, S_3, S_5, \ldots go down, and the gap between neighbors is one term b_n, which shrinks to 0. The two groups are squeezed onto the same number, the sum S.

What the test does not say. If condition (2) fails, the series diverges by the divergence test. If only condition (1) fails, the alternating series test gives no answer.

Error preview. Because the sum sits between any two neighboring partial sums, stopping after n terms makes an error no bigger than the next term, b_{n+1}. A later lesson, the alternating series error bound, uses this.

How to use the alternating series test

1. Write the series as \sum (-1)^{n+1} b_n or \sum (-1)^n b_n and identify b_n \gt 0.
2. Find \lim b_n. If it is not 0, the series diverges, and you are done.
3. Show b_n is decreasing: compare b_{n+1} with b_n, or show that the matching function has a negative derivative.
4. If both conditions hold, the series converges.

Alternating series test examples

Example 1: The alternating harmonic series

Does \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n} = 1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \cdots converge?

b_n = \dfrac{1}{n}

\lim_{n \to \infty} \dfrac{1}{n} = 0

n + 1 \gt n \quad \Rightarrow \quad \dfrac{1}{n + 1} \lt \dfrac{1}{n}

Answer: the series converges by the alternating series test. Its sum is ln 2 ≈ 0.693. Compare this with the harmonic series 1 + 1/2 + 1/3 + ⋯, which diverges: the alternating signs make the difference.

Example 2: Decreasing only from some point on

Does \sum_{n=1}^{\infty} \dfrac{(-1)^n n}{n^2 + 4} converge?

b_n = \dfrac{n}{n^2 + 4}

\lim_{n \to \infty} \dfrac{n}{n^2 + 4} = \lim_{n \to \infty} \dfrac{\frac{1}{n}}{1 + \frac{4}{n^2}} = 0

f(x) = \dfrac{x}{x^2 + 4}

f'(x) = \dfrac{(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = \dfrac{4 - x^2}{(x^2 + 4)^2}

4 - x^2 \lt 0 \text{ for } x \gt 2

Answer: the series converges. The sizes go 0.2, 0.25, 0.23, 0.2, …: they rise once, then fall for good, which is enough.

Example 3: When the terms do not shrink to 0

Does \sum_{n=1}^{\infty} \dfrac{(-1)^n 2n}{3n + 1} converge?

b_n = \dfrac{2n}{3n + 1}

\lim_{n \to \infty} \dfrac{2n}{3n + 1} = \lim_{n \to \infty} \dfrac{2}{3 + \frac{1}{n}} = \dfrac{2}{3}

\dfrac{2}{3} \neq 0

Answer: the series diverges by the divergence test. The alternating series test cannot be used here.

Example 4: A log over n

Does \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}\ln n}{n} converge?

b_n = \dfrac{\ln n}{n}

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

f'(x) = \dfrac{x \cdot \frac{1}{x} - \ln x}{x^2} = \dfrac{1 - \ln x}{x^2}

1 - \ln x \lt 0 \text{ for } x \gt e

Answer: the series converges by the alternating series test.

Example 5: A swinging toy (word problem)

A toy on a track moves right \dfrac{1}{\sqrt{n}} meters on its nth move when n is odd, and left by that amount when n is even. (a) Does its position settle at one point? (b) Is the total distance it travels finite?

x = 1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{4}} + \cdots

\lim_{n \to \infty} \dfrac{1}{\sqrt{n}} = 0

\sqrt{n + 1} \gt \sqrt{n} \quad \Rightarrow \quad \dfrac{1}{\sqrt{n + 1}} \lt \dfrac{1}{\sqrt{n}}

\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}

p = \tfrac{1}{2} \le 1

Answer: (a) yes, the position converges; it settles at about 0.60 m right of the start. (b) No, the total distance is infinite. The toy travels forever but ends up near one spot.

Common mistakes

Related methods

Absolute and conditional convergence

The alternating harmonic series converges, but the series of its sizes, \sum 1/n, diverges. A series like that is called conditionally convergent. If the series of sizes also converges, as for \sum (-1)^n/n^2, the series is absolutely convergent. That distinction is the next lesson, absolute and conditional convergence.

Estimating the sum

Since the error after n terms is at most b_{n+1}, you can choose how many terms to add. For the alternating harmonic series, 1000 terms give an error less than 0.001, since the next term is 1/1001.

Practice problems

1. Does \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2} converge?
2. Does \sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt{n}} converge?
3. Does \sum_{n=1}^{\infty} \dfrac{(-1)^n (n + 1)}{n} converge?
4. Does \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2n + 1} converge?
5. Does \sum_{n=1}^{\infty} \dfrac{(-1)^n n}{n^2 + 1} converge?
6. Does \sum_{n=1}^{\infty} \dfrac{\cos(n\pi)}{\sqrt[3]{n}} converge?
7. Does \sum_{n=1}^{\infty} \dfrac{(-1)^n n}{e^n} converge?
8. Does \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} n^2}{n^2 + 5} converge?
9. Does \sum_{n=2}^{\infty} \dfrac{(-1)^n}{\ln n} converge?
10. A thermostat makes corrections of +2, −1, +2/3, −1/2, … degrees, so the nth correction is \dfrac{2(-1)^{n+1}}{n}. Does the temperature change settle at one value? If so, what is it? (Use \sum (-1)^{n+1}/n = \ln 2.)

b_n = \dfrac{1}{n^2} \to 0

\dfrac{1}{(n + 1)^2} \lt \dfrac{1}{n^2}

b_n = \dfrac{1}{\sqrt{n}} \to 0

\dfrac{1}{\sqrt{n + 1}} \lt \dfrac{1}{\sqrt{n}}

b_n = \dfrac{n + 1}{n} = 1 + \dfrac{1}{n}

\lim_{n \to \infty} \left(1 + \dfrac{1}{n}\right) = 1 \neq 0

b_n = \dfrac{1}{2n + 1} \to 0

\dfrac{1}{2n + 3} \lt \dfrac{1}{2n + 1}

b_n = \dfrac{n}{n^2 + 1} \to 0

f'(x) = \dfrac{(x^2 + 1) - 2x^2}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}

1 - x^2 \le 0 \text{ for } x \ge 1

\cos(n\pi) = (-1)^n

b_n = \dfrac{1}{\sqrt[3]{n}} \to 0

\dfrac{1}{\sqrt[3]{n + 1}} \lt \dfrac{1}{\sqrt[3]{n}}

b_n = \dfrac{n}{e^n}

\lim_{x \to \infty} \dfrac{x}{e^x} = \lim_{x \to \infty} \dfrac{1}{e^x} = 0

f'(x) = \dfrac{e^x - xe^x}{e^{2x}} = \dfrac{1 - x}{e^x}

\dfrac{1 - x}{e^x} \le 0 \text{ for } x \ge 1

b_n = \dfrac{n^2}{n^2 + 5}

\lim_{n \to \infty} \dfrac{1}{1 + \frac{5}{n^2}} = 1 \neq 0

b_n = \dfrac{1}{\ln n} \to 0

\ln(n + 1) \gt \ln n \quad \Rightarrow \quad \dfrac{1}{\ln(n + 1)} \lt \dfrac{1}{\ln n}

\sum_{n=1}^{\infty} \dfrac{2(-1)^{n+1}}{n}

\sum_{n=1}^{\infty} \dfrac{2(-1)^{n+1}}{n} = 2\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}

2\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n} = 2\ln 2 \approx 1.386

Alternating series test FAQ

What are the two conditions of the alternating series test?

The sizes of the terms, b_n, must decrease (at least eventually), and they must approach 0. If both hold, the alternating series converges.

Can the alternating series test prove that a series diverges?

No. It only proves convergence. If the terms do not approach 0, use the divergence test to show divergence. If the terms are not decreasing, the test gives no answer.

Is the alternating series test the same as the Leibniz test?

Yes. The alternating series test is also called the Leibniz test, after Gottfried Leibniz, who used it.

Why does the alternating harmonic series converge when the harmonic series diverges?

The alternating signs make each term cancel part of the one before. The partial sums zigzag in a shrinking range around ln 2 instead of growing without bound.

Summary

• An alternating series \sum (-1)^{n+1} b_n with b_n \gt 0 converges if b_n decreases and b_n \to 0.
• Check the limit first: if b_n does not go to 0, the series diverges.
• Decreasing only has to hold from some point on; a derivative check often shows it.
• The partial sums zigzag around the sum, and the error after n terms is at most b_{n+1}.
• An alternating series can converge even when the series of its sizes diverges.

Previous lesson: Limit Comparison Test

Next lesson: Absolute and Conditional Convergence

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