Absolute and Conditional Convergence

A series with terms of both signs can converge because its terms are small, or only because positive and negative terms cancel. Comparing

\sum a_n

with the series of sizes

\sum |a_n|

separates the two cases.

Two kinds of convergence

Here's why the last statement holds. Each

a_n + |a_n|

is either 0 or

2|a_n|

, depending on the sign of

a_n

, so

0 \le a_n + |a_n| \le 2|a_n|

. If

\sum |a_n|

converges, then

\sum (a_n + |a_n|)

converges by direct comparison. Subtracting the convergent series

\sum |a_n|

leaves

\sum a_n

, which converges too.

Every series falls into exactly one of three classes: absolutely convergent, conditionally convergent, or divergent. For a series with positive terms, converging and converging absolutely are the same thing, so no such series converges conditionally.

Example 1

Decide whether each series converges absolutely, converges conditionally, or diverges.

(a)

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

(b)

\displaystyle\sum_{n=1}^{\infty} (-1)^n\,\frac{n + 1}{n^3 + 5}

(c)

\displaystyle\sum_{n=1}^{\infty} (-1)^n\,\frac{n}{\sqrt{n^2 + 4}}

(a) First test the sizes,

\frac{2n + 1}{n^2 + n}

. They behave like

\frac{2}{n}

, so compare with

\frac{1}{n}

\begin{aligned} &\lim_{n\to\infty} \frac{(2n + 1)/(n^2 + n)}{1/n} \\[4pt] &\qquad = \lim_{n\to\infty} \frac{2n^2 + n}{n^2 + n} = 2 \end{aligned}

Since

0 < 2 < \infty

and the harmonic series diverges,

\sum |a_n|

diverges by limit comparison. The series doesn't converge absolutely.

Now the alternating series test. Partial fractions give

\frac{2n + 1}{n(n + 1)} = \frac{1}{n} + \frac{1}{n + 1}

, a sum of two decreasing sequences, so the sizes decrease. Both pieces approach 0, so the sizes do too. The series converges by the alternating series test, so it converges conditionally.

(b) The sizes are

\frac{n + 1}{n^3 + 5}

, which behave like

\frac{1}{n^2}

\begin{aligned} &\lim_{n\to\infty} \frac{(n + 1)/(n^3 + 5)}{1/n^2} \\[4pt] &\qquad = \lim_{n\to\infty} \frac{n^3 + n^2}{n^3 + 5} = 1 \end{aligned}

Since

0 < 1 < \infty

and

\sum \frac{1}{n^2}

is a p-series with

p = 2 > 1

\sum |a_n|

converges by limit comparison. The series converges absolutely.

\frac{n}{\sqrt{n^2 + 4}} = \frac{1}{\sqrt{1 + 4/n^2}}

approach 1. The terms swing near

1

and

-1

and have no limit, so the series diverges by the nth term test.

Rearranging the terms

Adding finitely many numbers in a different order never changes the sum. For an absolutely convergent series that's still true: any regrouping or rearrangement of the terms has the same sum.

A conditionally convergent series is different. Take the alternating harmonic series, whose sum is

\ln 2

, and follow each positive term by the next four negative ones.

\begin{aligned} &1 - \tfrac{1}{2} - \tfrac{1}{4} - \tfrac{1}{6} - \tfrac{1}{8} \\[4pt] &\qquad + \tfrac{1}{3} - \tfrac{1}{10} - \tfrac{1}{12} - \tfrac{1}{14} - \tfrac{1}{16} + \cdots \end{aligned}

Every term of the original series appears exactly once. The partial sums after whole groups of five terms are

Terms	50	500	5000
Partial sum	$-0.0060$	$-0.0006$	$-0.0001$

and they close in on 0, far from

\ln 2 \approx 0.693

. With conditional convergence, the sum depends on the order of the terms.

Choosing a test

The tests overlap, and most series can be settled more than one way. Look at the form of the terms first.

What the terms look like	What to try
They don't approach 0	The nth term test: the series diverges.
$ar^n$ , or $\frac{1}{n^p}$	A geometric series converges if and only if $\|r\| < 1$ , and a p-series if and only if $p > 1$ .
Powers and roots of $n$ only	Limit comparison with a p-series, keeping the leading terms.
Factorials, or $n$ in an exponent	The ratio test.
Signs that alternate	Test $\sum \|a_n\|$ first. If the sizes don't approach 0, the series diverges. If the sizes approach 0 but $\sum \|a_n\|$ diverges, try the alternating series test.
$f(n)$ with an easy antiderivative	The integral test, after checking that $f$ is positive, continuous and decreasing.