What is the difference between the Alternating Series Test and the Ratio Test?

The Alternating Series Test only applies to series with alternating signs and determines convergence of the series itself. The Ratio Test works on any series and tests absolute convergence. If your series has alternating signs, apply the Alternating Series Test first; if you need absolute convergence or the series does not alternate, use the Ratio Test instead.

Does the Alternating Series Test tell me the sum of the series?

No, it only guarantees that a finite sum exists. To find the actual sum value, you need other methods such as recognizing a geometric series, using a Taylor series expansion, or computing partial sums. The test is purely a convergence/divergence result.

What does b_n \downarrow 0 mean?

The notation b_n \downarrow 0 means the sequence b_n is monotonically decreasing (each term is strictly smaller than the previous one) and the limit as n \to \infty is zero. Both the decreasing property and the zero limit are required; the arrow symbol compresses both conditions into one compact notation.

Can I use the Alternating Series Test on \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}?

Yes. Here b_n = 1/n^2, which is positive, strictly decreasing (since 1 > 1/4 > 1/9 > 1/16 > \cdots), and approaches zero. All conditions are satisfied, so the series converges by the Alternating Series Test. This is also an example of absolute convergence, since \sum 1/n^2 converges as a p-series.

Alternating Series Test

The Alternating Series Test determines whether a series with alternating signs converges when its terms decrease monotonically to zero.

\sum (-1)^n b_n \text{ converges if } b_n \downarrow 0

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What each symbol means

What Alternating Series Test takes

Symbol	Meaning
$b_n$	The sequence of positive terms in the series; each $b_n$ must be positive, decreasing, and approach zero for the test to apply, or the test tells you nothing about convergence.
$n$	The index or counter variable starting at 1 or 0, indicating which term of the series you are examining; do not confuse it with the coefficient $(-1)^n$ or with the value of each term itself.

When to use it

Use this test when you have a series with alternating signs and need to check if it converges.

Level

Usually taught in: Calculus II

Worked examples

1. Alternating harmonic series

Problem

Does the series

\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}

converge or diverge?

$b_n = \frac{1}{n}$
We identify $b_n$ as the positive part of each term by extracting the $(-1)^n$ factor and isolating $1/n$ .
$\frac{1}{1} > \frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \cdots$
We verify that $b_n$ is strictly decreasing because as $n$ grows, the denominator increases, making the fraction smaller.
$\lim_{n \to \infty} \frac{1}{n} = 0$
We check that the limit of $b_n$ is zero; as $n \to \infty$ , the reciprocal $1/n$ approaches zero.
$\text{Both conditions hold} \Rightarrow \sum_{n=1}^{\infty} (-1)^n \frac{1}{n} \text{ converges}$
Since $b_n$ is positive, decreasing, and approaches zero, the Alternating Series Test guarantees convergence.

Answer: $\sum_{n=1}^{\infty} (-1)^n \frac{1}{n} \text{ converges}$

This is the alternating harmonic series. Although the standard harmonic series $\sum 1/n$ diverges, the alternating version converges because the alternating signs create cancellation. The Alternating Series Test directly proves convergence without requiring us to compute the sum.

2. Series failing the limit condition

Problem

Does

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

converge?

$b_n = \frac{2n+1}{3n+2}$
We extract the positive part by identifying $(-1)^n$ for alternation and $b_n = \frac{2n+1}{3n+2}$ as the magnitude.
$\lim_{n \to \infty} \frac{2n+1}{3n+2} = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{3 + \frac{2}{n}} = \frac{2}{3}$
We compute the limit by dividing numerator and denominator by $n$ ; the limit is $2/3$ , not zero.
$b_n \not\to 0 \Rightarrow \text{Alternating Series Test fails}$
The Alternating Series Test requires $b_n \to 0$ , but this sequence approaches $2/3 \neq 0$ , so we cannot use this test.
$\text{By Divergence Test, if } b_n \not\to 0 \text{ then } \sum b_n \text{ diverges}$
The Divergence Test states that if the terms of any series do not approach zero, the series must diverge.

Answer: $\sum_{n=1}^{\infty} (-1)^n \frac{2n+1}{3n+2} \text{ diverges}$

The first necessary condition for any series to converge is that its terms must approach zero. This series fails that requirement, so it diverges regardless of alternating signs. The Divergence Test immediately resolves the question without needing to check other conditions.

3. Alternating profit in a coffee shop

Problem

A coffee shop's daily profit over many days has magnitude 100, 50, 25, 12.50, ... dollars, alternating between profit and loss. Does the cumulative profit converge to a finite value?

$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{100}{2^{n-1}}, \quad b_n = \frac{100}{2^{n-1}}$
The series has form $(-1)^{n+1}$ (positive on odd days, negative on even). We identify $b_n = \frac{100}{2^{n-1}}$ as the daily magnitude.
$b_1 = \frac{100}{2^0} = 100, \quad b_2 = \frac{100}{2^1} = 50, \quad b_3 = \frac{100}{2^2} = 25, \quad b_4 = \frac{100}{2^3} = 12.5$
The first few terms show a strictly decreasing sequence: each term is exactly half the previous term.
$\lim_{n \to \infty} \frac{100}{2^{n-1}} = 0$
As $n \to \infty$ , the denominator $2^{n-1}$ grows exponentially, so the fraction approaches zero.
$\text{Both } b_n \downarrow 0 \text{ and alternating} \Rightarrow \text{series converges}$
All conditions of the Alternating Series Test are satisfied: $b_n$ is positive, decreasing, and approaches zero.

Answer: $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{100}{2^{n-1}} \text{ converges to } \frac{200}{3}$

Real-world quantities like profit, error, or temperature change often follow alternating patterns that decay. The Alternating Series Test applies whenever magnitudes decrease monotonically to zero, guaranteeing a finite total. Here the cumulative profit approaches exactly $\frac{200}{3}$ dollars, even though individual days alternate and shrink without bound.

Common mistakes

Where Alternating Series Test usually goes wrong

The mistake	Why it is wrong	The fix
Checking only that $\lim b_n = 0$ and skipping the monotonicity condition.	A sequence can approach zero without being monotonically decreasing; it could oscillate, fluctuate unpredictably, or approach zero from above and below. The Alternating Series Test requires both $b_n \to 0$ AND $b_n$ strictly decreasing.	Always verify two separate facts: first that $b_n \to 0$ , and second that $b_1 > b_2 > b_3 > \cdots$ holds for all $n$ . Both conditions must hold simultaneously.
Concluding that the series converges absolutely after the Alternating Series Test succeeds.	The Alternating Series Test proves convergence of $\sum (-1)^n b_n$ itself, not that $\sum \|(-1)^n b_n\| = \sum b_n$ converges. A series can converge conditionally (alternating version converges) while failing absolute convergence (positive version diverges).	To test absolute convergence, separately analyze $\sum b_n$ using the Ratio Test, Integral Test, or Comparison Test. Do not assume the alternating version's convergence implies the positive version converges.
Applying the test to a series that is not alternating or misidentifying $b_n$ as including the sign.	The test is specific to alternating series of the form $\sum (-1)^n b_n$ where $b_n > 0$ . If the series lacks consistent alternation, or if you accidentally fold the sign into $b_n$ so that $b_n$ takes both positive and negative values, the test framework collapses.	Before applying the test, confirm the series has the form $\sum (-1)^n b_n$ or $\sum (-1)^{n+1} b_n$ and that $b_n$ is always positive. Separate the sign factor from the magnitude cleanly.

Tips and when to use something else

Check both conditions: $b_n$ must be strictly decreasing (not just eventually decreasing) and must approach zero. One without the other is insufficient.
The Alternating Series Test guarantees convergence of the series but not absolute convergence. If you need absolute convergence, apply the Ratio Test or Integral Test to $\sum b_n$ instead.
When $b_n$ fails to approach zero, apply the Divergence Test immediately—you do not need to check monotonicity or any other condition, and the series diverges.
Compare with the Ratio Test when dealing with factorials or exponential terms; the Ratio Test often detects convergence or divergence faster than verifying monotonicity by hand.

Related formulas

Geometric SeriesSums the first n terms of a geometric sequence—those where each term is the previous one times a constant—using a single formula instead of adding.Harmonic SeriesThe Harmonic Series, sum of reciprocals of positive integers, diverges infinitely—a foundational benchmark for determining whether other series converge.Ratio TestDetermine whether an infinite series converges by checking if the ratio of consecutive terms approaches a value less than one.Integral TestTest series convergence by checking improper integral ∫₁^∞ f(x)dx when f is positive, continuous, and decreasing; the series ∑f(n) shares convergence behavior.p-Series TestThe p-Series Test determines convergence of series 1/n^p: the series converges if and only if the exponent p is strictly greater than 1.

Frequently asked questions

What is the difference between the Alternating Series Test and the Ratio Test?: The Alternating Series Test only applies to series with alternating signs and determines convergence of the series itself. The Ratio Test works on any series and tests absolute convergence. If your series has alternating signs, apply the Alternating Series Test first; if you need absolute convergence or the series does not alternate, use the Ratio Test instead.
Does the Alternating Series Test tell me the sum of the series?: No, it only guarantees that a finite sum exists. To find the actual sum value, you need other methods such as recognizing a geometric series, using a Taylor series expansion, or computing partial sums. The test is purely a convergence/divergence result.
What does $b_n \downarrow 0$ mean?: The notation $b_n \downarrow 0$ means the sequence $b_n$ is monotonically decreasing (each term is strictly smaller than the previous one) and the limit as $n \to \infty$ is zero. Both the decreasing property and the zero limit are required; the arrow symbol compresses both conditions into one compact notation.
Can I use the Alternating Series Test on $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ ?: Yes. Here $b_n = 1/n^2$ , which is positive, strictly decreasing (since $1 > 1/4 > 1/9 > 1/16 > \cdots$ ), and approaches zero. All conditions are satisfied, so the series converges by the Alternating Series Test. This is also an example of absolute convergence, since $\sum 1/n^2$ converges as a $p$ -series.

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Reviewed 2026-09-18