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Alternating Series Test

Alternating Series Test

A convergence test for alternating series.

Consider the following alternating series (where an > 0 for all n) and/or its equivalents:

\[\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}{a_k}} = {a_1} - {a_2} + {a_3} - {a_4} + \cdots \]

The series converges if the following conditions are met.

1. an ≥ an +1 for all n ≥ N, where N ≥ 1, and

2. \(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)

 

See also

Convergent series, divergent series, limit, alternating series remainder

Key Formula

∑k=1∞(−1)k+1ak converges if: (1) an≥an+1 for all n≥N,(2) lim⁡n→∞an=0\sum_{k=1}^{\infty} (-1)^{k+1} a_k \text{ converges if: } \quad (1)\ a_n \geq a_{n+1} \text{ for all } n \geq N, \quad (2)\ \lim_{n \to \infty} a_n = 0
Where:
  • aka_k = The absolute value of the k-th term of the series, where a_k > 0 for all k
  • (−1)k+1(-1)^{k+1} = The alternating sign factor that makes consecutive terms switch between positive and negative
  • NN = A positive integer beyond which the decreasing condition must hold (it does not need to hold for all terms, just eventually)

Worked Example

Problem: Determine whether the alternating harmonic series converges: ∑k=1∞(−1)k+1k=1−12+13−14+⋯\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
Step 1: Identify the sequence a_k. Here, the terms without the alternating sign are:
ak=1ka_k = \frac{1}{k}
Step 2: Check condition 1: Is the sequence decreasing? Compare consecutive terms:
an=1n≥1n+1=an+1for all n≥1a_n = \frac{1}{n} \geq \frac{1}{n+1} = a_{n+1} \quad \text{for all } n \geq 1
Step 3: This inequality holds because a larger denominator produces a smaller fraction. So the sequence is decreasing.
Step 4: Check condition 2: Does the sequence approach zero?
lim⁡n→∞1n=0✓\lim_{n \to \infty} \frac{1}{n} = 0 \quad \checkmark
Step 5: Both conditions are satisfied, so the Alternating Series Test confirms convergence.
Answer: The alternating harmonic series converges by the Alternating Series Test. (Note: the ordinary harmonic series diverges, so this is a case of conditional convergence.)

Another Example

This example shows what happens when one of the two conditions fails. It reinforces that the test has requirements that must both be checked, and that failing the limit condition means the series diverges outright.

Problem: Determine whether the series converges: ∑k=1∞(−1)k+1⋅kk+1\sum_{k=1}^{\infty} \frac{(-1)^{k+1} \cdot k}{k+1}
Step 1: Identify the sequence a_k (the positive part of each term):
ak=kk+1a_k = \frac{k}{k+1}
Step 2: Check condition 2 first: Does the sequence approach zero?
lim⁡n→∞nn+1=lim⁡n→∞11+1n=1≠0\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1 \neq 0
Step 3: Since the limit is not zero, condition 2 fails. The Alternating Series Test does not apply.
Step 4: Moreover, because the terms do not approach zero, the series actually diverges by the Divergence Test (nth-Term Test).
lim⁡n→∞(−1)n+1⋅nn+1≠0  ⟹  series diverges\lim_{n \to \infty} \frac{(-1)^{n+1} \cdot n}{n+1} \neq 0 \implies \text{series diverges}
Answer: The series diverges. The Alternating Series Test cannot be applied because the limit of a_n is not zero.

Frequently Asked Questions

What happens if the terms are not decreasing but the limit is zero?
The Alternating Series Test requires both conditions. If the terms are not eventually decreasing, the test is inconclusive — it does not tell you whether the series converges or diverges. You would need a different test. However, note that the decreasing condition only needs to hold eventually (for all n beyond some threshold N), not necessarily from the very first term.
Does the Alternating Series Test prove absolute convergence or conditional convergence?
The Alternating Series Test only proves that the series converges; it does not distinguish between absolute and conditional convergence. To determine absolute convergence, you must separately test whether the series of absolute values converges. If the absolute-value series diverges but the alternating series converges, the convergence is conditional.
When should you use the Alternating Series Test instead of the Ratio Test?
Use the Alternating Series Test when the series has terms that alternate in sign and you can verify the two conditions directly. The Ratio Test works on any series (alternating or not), but it is often inconclusive for series like the alternating harmonic series where the ratio of consecutive terms approaches 1. The Alternating Series Test is specifically designed for sign-changing series and succeeds in many cases where the Ratio Test fails.

Alternating Series Test vs. Ratio Test

Alternating Series TestRatio Test
What it testsConvergence of alternating series specificallyConvergence (including absolute) of any series
Key conditionTerms must be eventually decreasing and approach zeroThe limit of |a_{n+1}/a_n| must be less than 1
When inconclusiveWhen the decreasing condition fails (limit = 0 alone is not enough)When the limit of the ratio equals exactly 1
Type of convergence provedConvergence (may be conditional)Absolute convergence (when it concludes convergence)
Best used forSeries like (-1)^n / n^p, (-1)^n / ln(n)Series involving factorials, exponentials, or nth powers

Why It Matters

The Alternating Series Test appears throughout AP Calculus BC and college-level Calculus II courses as one of the essential convergence tests students must master. It is often the only test that works for series like the alternating harmonic series, where more general tests (Ratio, Root) are inconclusive. Understanding this test is also the foundation for the Alternating Series Remainder (error bound), which lets you estimate how close a partial sum is to the true value of the series.

Common Mistakes

Mistake: Assuming the series diverges just because the Alternating Series Test conditions are not met.
Correction: Failing the test only means the test is inconclusive (unless the limit of a_n ≠ 0, which guarantees divergence by the Divergence Test). If only the decreasing condition fails, the series might still converge — you need a different method to decide.
Mistake: Forgetting to check that the sequence is decreasing and only verifying that the limit equals zero.
Correction: Both conditions are required. A series can have terms approaching zero without being decreasing, and in some such cases the series diverges. Always verify both conditions. You can check the decreasing condition by showing a_{n+1} ≤ a_n, or equivalently by showing that the function f(x) corresponding to a_n has a negative derivative for large x.

Related Terms