Math Core

The alternating harmonic series converges, but only because its positive and negative terms cancel. The alternating series $\sum \frac{(-1)^n}{n^2}$ converges for a much sturdier reason: it would converge even if every term were positive. This lesson gives names to these two kinds of convergence and shows you how to classify any series with mixed signs.

Two kinds of convergence

For any series $\sum a_n$ , you can form the series of absolute values, $\sum |a_n|$ , which has all nonnegative terms.

Definition

Absolute and conditional convergence

$\sum a_n$ converges absolutely if $\sum |a_n|$ converges.
$\sum a_n$ converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.

The key fact connecting them is that absolute convergence is the stronger property.

Absolute convergence implies convergence

If $\sum |a_n|$ converges, then $\sum a_n$ converges.

Why? Making some terms negative can only help the partial sums stay bounded; it can't create divergence. (Formally, $0 \le a_n + |a_n| \le 2|a_n|$ , so $\sum (a_n + |a_n|)$ converges by comparison, and subtracting the convergent $\sum |a_n|$ leaves $\sum a_n$ convergent.)

This theorem is especially useful for series whose signs change irregularly, like $\sum \frac{\sin n}{n^2}$ . These are not alternating, so the alternating series test doesn't apply, but you can often show that the absolute values converge.

A classification procedure

Every series falls into exactly one of three categories: absolutely convergent, conditionally convergent, or divergent. Here is a reliable order of attack.

Check the terms. If $a_n \not\to 0$ , the series diverges. Stop.
Test the absolute values. Apply a positive-term test (p-series, comparison, ratio, integral) to $\sum |a_n|$ . If it converges, the series converges absolutely. Stop.
If $\sum |a_n|$ diverges, test the original series. For an alternating series, use the alternating series test. If it converges, the series converges conditionally.

Common mistake

The divergence of $\sum |a_n|$ does not mean $\sum a_n$ diverges. For example, $\sum \frac{1}{n}$ diverges, yet $\sum \frac{(-1)^{n+1}}{n}$ converges. If the absolute series diverges, you still have to test the original series before you decide. (The one exception is the ratio test: if $L > 1$ , the original series diverges too, because its terms don't go to $0$ .)

Worked examples

Worked example: Absolutely convergent

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$ .

Solution. The absolute values form $\sum \frac{1}{n^3}$ , a p-series with $p = 3 > 1$ , which converges. So the series converges absolutely.

Worked example: Conditionally convergent

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n + 1}$ .

Solution. Absolute values: $\sum \frac{1}{2n+1}$ . Limit comparison with $\sum \frac1n$ gives $\dfrac{n}{2n+1} \to \dfrac12$ , positive and finite, so $\sum \frac{1}{2n+1}$ diverges.

Original series: $\frac{1}{2n+1}$ decreases and approaches $0$ , so the alternating series test says the series converges.

It converges, but not absolutely: the series converges conditionally.

Worked example: Irregular signs

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ .

Solution. The signs of $\cos n$ follow no regular pattern, so this is not an alternating series. Test absolute values instead: $0 \le \dfrac{|\cos n|}{n^2} \le \dfrac{1}{n^2}$ , and $\sum \frac{1}{n^2}$ converges. By direct comparison, $\sum \left|\frac{\cos n}{n^2}\right|$ converges, so the series converges absolutely.

Worked example: Using the ratio test

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-3)^n}{n!}$ .

Solution. Apply the ratio test to the absolute values:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1} \to 0.

Since $L = 0 < 1$ , the series converges absolutely.

Why the distinction matters

Absolutely convergent series behave like finite sums: you can rearrange their terms in any order and the sum stays the same. Conditionally convergent series are fragile. A famous theorem of Riemann says that the terms of a conditionally convergent series can be rearranged to add up to any number you like, or even to diverge. For instance, taking the terms of $1 - \frac12 + \frac13 - \frac14 + \cdots$ in the order "two positives, then one negative" produces a different sum ( $\frac32 \ln 2$ instead of $\ln 2$ ). The distinction also shows up at the endpoints of intervals of convergence for power series, later in this unit.

Tip

A good first question for any series with negative terms: "Does $\sum |a_n|$ converge?" If yes, you are done, and the answer is the strongest possible: absolute convergence.

Practice

Practice 1

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt n}$ .

Practice 2

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ .

Practice 3

Classify $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{n + 3}$ .

Practice 4

Classify $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 1}$ .

Practice 5

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{\sin n}{n\sqrt{n}}$ .

Practice 6

For which values of $p$ is $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ conditionally convergent? Write a compound inequality in $p$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 7

Which of the following series converges conditionally?