The Alternating Series Test
The alternating series test in two conditions, worked examples, what it can't tell you about absolute convergence, and the alternating series error bound.
The Alternating Series Test
Most students assume that if the terms of an alternating series approach zero, the series converges. That is wrong. The alternating series test requires two conditions: the absolute values of the terms must decrease monotonically to zero. Without the decreasing condition, the series can diverge even when the limit is zero. The alternating series test covers series of the form Σ(-1)n-1bn or Σ(-1)nbn where bn > 0. For convergence under the alternating series test, bn must be decreasing for all n (or at least eventually) and limit bn = 0 must hold.
Statement: b<sub>n</sub> Decreasing and Tends to Zero
The formal statement from Stewart, Calculus, section 11.5, uses the series Σ(-1)n-1bn with bn > 0. Two conditions: (i) bn+1 ≤ bn for all n, (ii) limn→∞ bn = 0. When both hold, the alternating series converges. The theorem does not give the sum, only convergence. The College Board AP Calculus BC CED Topic 10.7 lists the alternating series test as a required technique for the BC exam. The same criteria appear in OpenStax Volume 2, section 5.2.
If either condition fails, the test is inconclusive. For example, Σ(-1)n(1 + 1/n) has terms that do not approach zero, so the divergence test applies instead. If bn decreases but not to zero, the series diverges by the nth-term test. If bn approaches zero but fails to decrease, the alternating series test says nothing, you must use another method.
Worked Examples
Example 1: Alternating Harmonic Series
Consider Σ(-1)n-1/n with n starting at 1. Here bn = 1/n. Check the conditions: 1/(n+1) ≤ 1/n for all n, so bn decreases. Limit bn = 0. Both hold, so the alternating harmonic series converges. Its sum is ln 2, but the alternating series test only guarantees convergence, not the sum value.
Example 2: Series That Fails the Decreasing Condition
Test Σ(-1)n (1/√n + (-1)n/n). The absolute terms do not decrease monotonically. The alternating series test does not apply. Use the ratio test or another test instead, the divergence test shows the limit of terms is not zero, so the series diverges.
Example 3: Series Satisfying the Test
Take Σ(-1)n / (n2+1). bn = 1/(n2+1) decreases for n≥1 and tends to zero. The alternating series test confirms convergence. The series converges conditionally, not absolutely, because Σ 1/(n2+1) converges by comparison to a p-series with p=2.
Alternating Harmonic Series
The alternating harmonic series, Σ(-1)n-1/n, is the classic example taught in Stewart section 11.5. It converges conditionally. The harmonic series itself (Σ 1/n) diverges, but alternating the signs causes cancellation that brings the partial sums to a finite limit. The College Board AP BC CED Topic 10.7 uses this series as the model for the alternating series test. The error bound applied to this series gives |RN| ≤ 1/(N+1).
Alternating Series Error Bound
The alternating series error bound is a separate result, covered as AP BC CED Topic 10.12. If a series converges by the alternating series test, the remainder after N terms satisfies |RN| ≤ bN+1. This means the error from using the Nth partial sum is at most the absolute value of the first omitted term. The bound is simple to apply: just take the next term. For the alternating harmonic series, using 10 terms gives |R10| ≤ 1/11 ≈ 0.0909.
Error Bound Example
Estimate Σ(-1)n-1/(n2) using the third partial sum. S3 = 1-1/4 + 1/9 = 0.8611. The fourth term is 1/16 = 0.0625. The true sum lies between S3 and S3 + 0.0625. More precisely, |R3| ≤ 0.0625. This bound holds because the series converges by the alternating series test (bn = 1/n2 decreases to zero).
The alternating series estimation theorem gives the same bound. Note that the bound applies only to alternating series that meet the test conditions, not to all series.
Then Test Absolute Convergence
After applying the alternating series test, always check absolute convergence separately. Take the absolute value of each term and apply a convergence test such as the p-series test or comparison test. If Σ |an| converges, the original series converges absolutely. If Σ |an| diverges but the alternating series converges conditionally. For the alternating harmonic series, Σ 1/n diverges (p=1), so the series converges conditionally, not absolutely. Only alternating series can exhibit conditional convergence; non-alternating series that converge do so absolutely.
Common Questions
What is the Leibniz test?
The Leibniz test is another name for the alternating series test. It names Gottfried Leibniz, who first stated the conditions for convergence of alternating series. The terms are the same: b<sub>n</sub> must decrease to zero.
Does the alternating series test give the sum?
No. The test only tells you whether the series converges, not what it sums to. For the alternating harmonic series, the sum is known to be ln 2, but that requires additional calculation, not the test.
What if b<sub>n</sub> decreases but limit is not zero?
Then the alternating series diverges by the nth-term test (divergence test). The alternating series test requires both conditions; failure of either means the test does not apply.
Can I use the alternating series error bound for any series?
No. The bound |R<sub>N</sub>| ≤ b<sub>N+1</sub> applies only to series that satisfy the alternating series test conditions. For other series, use the integral test remainder estimate or another method.
How do I know if b<sub>n</sub> is decreasing?
Check that b<sub>n+1</sub> ≤ b<sub>n</sub> for all n. For simple functions like 1/n or 1/(n<sup>2</sup>+1), you can show it by algebra. For more complex terms, consider the derivative of the continuous function f(x) corresponding to b<sub>n</sub>; if f'(x) ≤ 0, the sequence is decreasing.