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Alternating Series
Test for Convergence

Master the conditions that determine when an alternating series converges, and learn to apply the test with confidence on the AP exam.

AP Calculus BC

What Is an Alternating Series?

An alternating series is any infinite series whose terms switch between positive and negative. The general forms are:

The factor \((-1)^n\) or \((-1)^{n+1}\) is what causes the sign alternation. The sequence \(\{b_n\}\) represents the magnitude of each term.

Partial Sums Oscillate Toward a Limit

n (number of terms) Partial Sum L 1 2 3 4 5 6 7 8 9 10

Why the AST Matters

Convergence without Absolute
Many series converge only because their signs alternate: the AST detects this conditional convergence.
Error Estimation
The AST gives a built-in error bound: the error after \(n\) terms is at most \(b_{n+1}\).
AP Exam Staple
The test appears in free-response and multiple-choice questions nearly every year on the BC exam.

The Alternating Series Test

Both conditions focus on the magnitude sequence \(\{b_n\}\), not the alternating signs. You need both: decreasing alone is not enough, and approaching zero alone is not enough.

The Two Conditions Visualized

n \(b_n\) Decreasing: \(b_{n+1} \leq b_n\) → 0 1 2 3 4 5 6 7 8 9 10

Conditional vs. Absolute Convergence

Absolute Convergence
\(\sum |a_n|\) converges. This is stronger: if a series converges absolutely, it converges (period).
Conditional Convergence
\(\sum a_n\) converges but \(\sum |a_n|\) diverges. The alternating signs are essential for convergence.

What the AST Cannot Do

The Alternating Series Test can only confirm convergence. If either condition fails, the test is inconclusive: the series might still converge by another test, or it might diverge.

If \(\lim b_n \neq 0\)
The series diverges by the Divergence Test (nth-term test). The AST is not needed here.
If \(b_n\) is not decreasing
The AST is inconclusive. Try another test: Ratio, Root, or comparison tests may work.

How to Apply the AST

1
Identify the series as alternating. Confirm the factor \((-1)^n\) or \((-1)^{n+1}\) is present, and isolate \(b_n > 0\).
2
Check: \(\lim_{n\to\infty} b_n = 0\). If this fails, the series diverges by the Divergence Test: stop here.
3
Check: \(b_{n+1} \leq b_n\) eventually. Show the magnitudes are decreasing. You can test \(\frac{b_{n+1}}{b_n} \leq 1\), or show \(f'(x) < 0\) for \(f(x) = b_x\).
4
Conclude. If both conditions hold, the series converges by the Alternating Series Test.

Example 1: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\)

Goal: Determine whether the series converges using the AST.

1
Identify: Alternating factor is \((-1)^{n+1}\). So \(b_n = \frac{1}{n^2}\).
2
Limit: \(\displaystyle\lim_{n\to\infty}\frac{1}{n^2} = 0\) ✓
3
Decreasing: \(\frac{1}{(n+1)^2} < \frac{1}{n^2}\) for all \(n \geq 1\) ✓
4
Conclusion: The series converges by the AST.

Example 2: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}\,n}{n+1}\)

Goal: Determine whether the series converges.

1
Identify: Alternating factor is \((-1)^n\). So \(b_n = \frac{n}{n+1}\).
2
Limit: \(\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1 \neq 0\) ✕

Example 3: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\ln n}{n}\)

Goal: Determine convergence. Here \(b_n = \frac{\ln n}{n}\).

1
Limit: \(\displaystyle\lim_{n\to\infty}\frac{\ln n}{n} \stackrel{L'H}{=} \lim_{n\to\infty}\frac{1/n}{1} = 0\) ✓
2
Decreasing: Let \(f(x) = \frac{\ln x}{x}\). Then \(f'(x) = \frac{1 - \ln x}{x^2}\). For \(x \geq 3\), \(\ln x > 1\), so \(f'(x) < 0\). Thus \(b_n\) is eventually decreasing. ✓
3
Conclusion: The series converges by the AST.

The Error Bound (Remainder Estimate)

Example: Approximate \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\) using 4 terms. The error is at most \(b_5 = \frac{1}{25} = 0.04\). The approximation \(S_4 = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} \approx 0.7986\) is within 0.04 of the true sum.

Example 4: How Many Terms?

Problem: How many terms of \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}\) are needed so the error is less than \(0.001\)?

1
Set up: We need \(b_{N+1} < 0.001\), i.e., \(\frac{1}{(N+1)^3} < 0.001\).
2
Solve: \((N+1)^3 > 1000\), so \(N+1 > 10\), giving \(N > 9\).
3
Answer: We need at least \(N = 10\) terms for the error to be less than \(0.001\).

Common Mistakes to Avoid

✕ Forgetting to check the limit
Students assume that if \(b_n\) is decreasing, the series converges. Both conditions are required.
✕ Testing \(a_n\) instead of \(b_n\)
The decreasing test applies to the magnitude \(b_n > 0\), not the alternating term \(a_n = (-1)^n b_n\).
✕ Claiming divergence when AST fails
If \(b_n\) is not decreasing but \(\lim b_n = 0\), the AST is inconclusive, not proof of divergence.

Convergence Tests Comparison

AST Ratio Test p-Series Test
Applies to Alternating series only Any series \(\sum \frac{1}{n^p}\) only
What you check \(b_n \downarrow\) and \(b_n \to 0\) \(\lim \frac{|a_{n+1}|}{|a_n|}\) Value of \(p\)
Determines absolute convergence? No Yes Yes
Error bound? Yes (\(\leq b_{N+1}\)) No No

Question 1 of 5

Which of the following is a necessary condition for the Alternating Series Test to confirm convergence?

Question 2 of 5

Does the series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n+3}\) converge by the AST?

Question 3 of 5

The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\) is an example of:

Question 4 of 5

Using the alternating series error bound, if \(S_5 = \sum_{n=1}^{5}\frac{(-1)^{n+1}}{n!}\), what is the maximum error of this approximation?

Question 5 of 5

Which of the following alternating series diverges?