Absolute and Conditional Convergence

Part 1 · Hook

Why this matters

1 - \frac{1}{2} + \frac{1}{3} - \cdots

converges, but strip the signs and you get the harmonic series, which diverges. Some series converge because their terms are small; others converge only because their signs keep cancelling. Telling the two apart is what absolute and conditional convergence are about.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converge?

Yes, by the alternating series test
No, since $\sum \frac{1}{n}$ diverges
No, by the nth term test
Yes, by the ratio test

Show the answer

The sizes $\frac{1}{n}$ decrease and approach 0, and the signs alternate.

Correct: Yes, by the alternating series test:
No, since $\sum \frac{1}{n}$ diverges:
No, by the nth term test:
Yes, by the ratio test:

2. For which $p$ does $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converge?

$p > 1$
$p \ge 1$
$p > 0$
$p < 1$

Show the answer

A $p$ -series converges exactly when $p > 1$ .

Correct: $p > 1$ :
$p \ge 1$ :
$p > 0$ :
$p < 1$ :

Part 4 · See it

See it first

Partial sums for N = 1 to 20. Open dots: the harmonic series, the sum of 1 over n, climbing steadily to about 3.60 with no ceiling. Filled dots: the alternating harmonic series, zigzagging around a dashed line at about 0.693. The alternating series converges but the series of absolute values diverges: conditional convergence. — Same sizes, different signs: $\sum \frac{1}{n}$ climbs without bound, while $\sum \frac{(-1)^{n+1}}{n}$ settles near $0.693$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Take the absolute value of every term $\sum |a_n|$ is a series of positive terms, which the comparison, integral and ratio tests can decide
$\sum |a_n|$ converges $\sum a_n$ converges too (absolutely): signs can only cancel, never add
$\sum |a_n|$ diverges but $\sum a_n$ convergesthe convergence comes from the signs cancelling: conditional convergence
$\sum a_n$ converges absolutelyits terms can be rearranged in any order without changing the sum

Part 6 · Key ideas

Key ideas

Absolutely convergent: $\sum |a_n|$ converges. Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ diverges.
Absolute convergence implies convergence.
To classify: test $\sum |a_n|$ first. If it converges, you are done (absolute). If not, test $\sum a_n$ itself, usually with the alternating series test.
A divergent $\sum |a_n|$ alone decides nothing about $\sum a_n$ .

Part 7 · Misconception

A common mistake

The wrong idea: $\sum \frac{1}{\sqrt{n}}$ diverges, so $\sum \frac{(-1)^n}{\sqrt{n}}$ diverges too.

What actually happens: A divergent series of absolute values only rules out absolute convergence. The signed series converges by the alternating series test, so it converges conditionally.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Four series with alternating signs

Four series, each with terms that alternate in sign
Series	Terms	Starts at
I	$\frac{(-1)^n}{n^2}$	$n = 1$
II	$\frac{(-1)^n}{\sqrt{n}}$	$n = 1$
III	$(-1)^n \frac{n}{n + 1}$	$n = 1$
IV	$\frac{(-1)^n}{n \ln n}$	$n = 2$

1. How does series I, $\sum \frac{(-1)^n}{n^2}$ , behave?

It converges conditionally: it alternates and $\frac{1}{n^2} \to 0$
It converges absolutely: $\sum \frac{1}{n^2}$ converges
It diverges: $\frac{1}{n^2}$ shrinks too slowly
It converges absolutely: its terms alternate in sign

Show the answer

$\sum \left|\frac{(-1)^n}{n^2}\right| = \sum \frac{1}{n^2}$ is a $p$ -series with $p = 2 > 1$ , so it converges. The series converges absolutely (and so it converges).

It converges conditionally: it alternates and $\frac{1}{n^2} \to 0$ : Conditional means $\sum |a_n|$ diverges, but $\sum \frac{1}{n^2}$ converges.
Correct: It converges absolutely: $\sum \frac{1}{n^2}$ converges: Right: the series of absolute values is a $p$ -series with $p = 2$ .
It diverges: $\frac{1}{n^2}$ shrinks too slowly: $\frac{1}{n^2}$ shrinks fast enough: $p = 2 > 1$ .
It converges absolutely: its terms alternate in sign: Alternating signs do not make convergence absolute; the absolute values must form a convergent series.

2. How does series II, $\sum \frac{(-1)^n}{\sqrt{n}}$ , behave?

It converges absolutely
It diverges, since $\sum \frac{1}{\sqrt{n}}$ diverges
It converges conditionally
It diverges by the nth term test

Show the answer

The signs alternate and $\frac{1}{\sqrt{n}}$ decreases to 0, so the series converges by the alternating series test. But $\sum \frac{1}{\sqrt{n}}$ diverges ( $p = \frac{1}{2} \le 1$ ). It converges conditionally.

It converges absolutely: $\sum \frac{1}{\sqrt{n}}$ is a $p$ -series with $p = \frac{1}{2}$ , which diverges.
It diverges, since $\sum \frac{1}{\sqrt{n}}$ diverges: The series of absolute values diverging does not make the signed series diverge.
Correct: It converges conditionally: Right: it converges by the alternating series test, but $\sum \frac{1}{\sqrt{n}}$ diverges.
It diverges by the nth term test: $\frac{1}{\sqrt{n}} \to 0$ , so the nth term test says nothing.

3. How does series III, $\sum (-1)^n \frac{n}{n + 1}$ , behave?

It converges conditionally, by the alternating series test
It converges absolutely, since $\frac{n}{n+1} < 1$
It converges conditionally, since $\sum \frac{n}{n+1}$ diverges
It diverges: its terms do not approach 0

Show the answer

$\frac{n}{n + 1} \to 1$ , so the terms swing near $\pm 1$ and do not approach 0. The series diverges by the nth term test.

It converges conditionally, by the alternating series test: The sizes approach 1, not 0, so that test does not apply.
It converges absolutely, since $\frac{n}{n+1} < 1$ : $\sum \frac{n}{n + 1}$ diverges; each term near 1 adds about 1.
It converges conditionally, since $\sum \frac{n}{n+1}$ diverges: A divergent series of absolute values is half the story; the series itself must converge, and it does not.
Correct: It diverges: its terms do not approach 0: Right: the sizes approach 1.

Table

A series and its absolute values

The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n + 1}$ and the series of its absolute values
Series	Written out
$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n + 1}$	$\frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \cdots$
$\sum_{n=1}^{\infty} \frac{1}{2n + 1}$	$\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \cdots$

4. How should $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n + 1}$ be classified?

Conditionally convergent
Absolutely convergent
Divergent
Both absolutely and conditionally convergent

Show the answer

The series converges (alternating series test) and $\sum |a_n| = \sum \frac{1}{2n + 1}$ diverges (limit comparison with $\frac{1}{n}$ ), so it is conditionally convergent.

Correct: Conditionally convergent: Right: it converges, but its absolute values do not.
Absolutely convergent: $\sum \frac{1}{2n + 1}$ diverges, so the convergence is not absolute.
Divergent: The alternating series test shows it converges.
Both absolutely and conditionally convergent: The two labels exclude each other: absolute means $\sum |a_n|$ converges, conditional means it diverges.

5. The signs of $\sin n$ follow no regular pattern. What can you conclude about $\sum_{n=1}^{\infty} \frac{\sin n}{n^2}$ ?

It converges by the alternating series test
Nothing, since the signs are irregular
It diverges, since $\sin n$ does not approach 0
It converges absolutely, since $\left|\frac{\sin n}{n^2}\right| \le \frac{1}{n^2}$

Show the answer

$0 \le \left|\frac{\sin n}{n^2}\right| \le \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, so $\sum \left|\frac{\sin n}{n^2}\right|$ converges by direct comparison. The series converges absolutely, so it converges.

It converges by the alternating series test: The signs do not alternate in a regular pattern, so the test does not apply.
Nothing, since the signs are irregular: Absolute convergence handles any pattern of signs.
It diverges, since $\sin n$ does not approach 0: The term is $\frac{\sin n}{n^2}$ , which does approach 0.
Correct: It converges absolutely, since $\left|\frac{\sin n}{n^2}\right| \le \frac{1}{n^2}$ : Right: compare the absolute values with a convergent $p$ -series.

6. You know that $\sum |a_n|$ diverges. What can you conclude about $\sum a_n$ ?

It diverges
Nothing yet: it may converge conditionally or diverge
It converges conditionally
It converges absolutely

Show the answer

A divergent $\sum |a_n|$ rules out absolute convergence and nothing more. $\sum \frac{(-1)^n}{n}$ converges and $\sum (-1)^n$ diverges, though both have divergent absolute values.

It diverges: The alternating harmonic series converges though $\sum \frac{1}{n}$ diverges.
Correct: Nothing yet: it may converge conditionally or diverge: Right: both happen, so test $\sum a_n$ itself.
It converges conditionally: $\sum (-1)^n$ has $\sum |a_n|$ divergent and itself diverges.
It converges absolutely: Absolute convergence needs $\sum |a_n|$ to converge.

Part 9 · Summary

Summary

Every series is absolutely convergent (

\sum |a_n|

converges), conditionally convergent (

\sum a_n

converges but

\sum |a_n|

diverges), or divergent. Absolute convergence implies convergence, and an absolutely convergent series keeps its sum when its terms are rearranged.

Part 10 · Up next

What comes next

Part 11 · Connections