Absolute Convergence

Q: Does absolute convergence imply convergence?

Yes. If $\sum |a_n|$ converges, then $\sum a_n$ also converges. This is a theorem, not a definition — the proof relies on the comparison test. The converse is false: a series can converge without converging absolutely (conditional convergence).

Q: How do you test for absolute convergence?

Replace every term $a_n$ with $|a_n|$ and test whether the resulting non-negative series converges. You can use any standard convergence test on $\sum |a_n|$, such as the ratio test, root test, comparison test, or p-series test. The ratio test and root test are especially convenient because they automatically test absolute convergence.

Absolute Convergence
Absolutely Convergent

Describes a series that converges when all terms are replaced by their absolute values. To see if a series converges absolutely, replace any subtraction in the series with addition. If the new series converges, then the original series converges absolutely.

Note: Any series that converges absolutely is itself convergent.

Definition:	The series $\sum\limits_{n = 1}^\infty {{a_n}} $ is absolutely convergent if the series $\sum\limits_{n = 1}^\infty {\left\| {{a_n}} \right\|} $ converges.

Example:	Determine if $\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots $ is absolutely convergent.

Solution:	To find out, consider the series \[\sum\limits_{n = 1}^\infty {\left\| {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \right\|} = \sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]
	This is an infinite geometric series with ratio $r = \frac{1}{2}$, so we know that it converges as shown. \[\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{{1 - \frac{1}{2}}} = 2\] As a result, we can conclude that $\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} $ converges absolutely.

Note:	The series $\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots $ and $\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots $ converge to different sums. In fact, $\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} $ is an infinite geometric series with ratio $r = - \frac{1}{2}$, so \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots = \frac{1}{{1 - \left( { - \frac{1}{2}} \right)}} = \frac{2}{3}\]

See also

Convergence tests

Key Formula

\sum_{n=1}^{\infty} a_n \text{ is absolutely convergent if } \sum_{n=1}^{\infty} |a_n| \text{ converges.}

Where:

$a_n$ = The general term of the series, which may be positive, negative, or alternating in sign
$|a_n|$ = The absolute value of each term, making every term non-negative
$n$ = The index of summation, typically starting at 1 and increasing to infinity

Worked Example

Problem: Determine whether the series

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

converges absolutely.

Step 1: Form the absolute value series by taking the absolute value of each term.

\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^2}\right| = \sum_{n=1}^{\infty} \frac{1}{n^2}

Step 2: Recognize the resulting series. This is a p-series with p = 2.

\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges when } p > 1

Step 3: Since p = 2 > 1, the p-series converges. In fact, it converges to a known value.

\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449

Step 4: Because the absolute value series converges, the original series converges absolutely.

\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \text{ is absolutely convergent.}

Answer: The series

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

converges absolutely because the p-series

\sum_{n=1}^{\infty} \frac{1}{n^2}

converges.

Another Example

This example shows a series that converges but fails absolute convergence, illustrating the important distinction between absolute and conditional convergence.

Problem: Determine whether the alternating harmonic series

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots

converges absolutely.

Step 1: Form the absolute value series by removing the alternating sign.

\sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}

Step 2: Identify the resulting series. This is the harmonic series, a p-series with p = 1.

\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

Step 3: The harmonic series diverges (since p = 1 is not greater than 1). So the series does NOT converge absolutely.

\sum_{n=1}^{\infty} \frac{1}{n} \text{ diverges}

Step 4: However, the original alternating series does converge by the Alternating Series Test (terms decrease to zero). This means the series is conditionally convergent, not absolutely convergent.

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2 \approx 0.6931

Answer: The alternating harmonic series is NOT absolutely convergent. It is conditionally convergent — it converges on its own, but its absolute value series diverges.

Frequently Asked Questions

What is the difference between absolute convergence and conditional convergence?

A series is absolutely convergent when the series of absolute values converges. A series is conditionally convergent when the original series converges but the series of absolute values diverges. Absolute convergence is the stronger condition: every absolutely convergent series is convergent, but not every convergent series is absolutely convergent. The alternating harmonic series is the classic example of conditional convergence.

Does absolute convergence imply convergence?

Yes. If

\sum |a_n|

converges, then

\sum a_n

also converges. This is a theorem, not a definition — the proof relies on the comparison test. The converse is false: a series can converge without converging absolutely (conditional convergence).

How do you test for absolute convergence?

Replace every term

a_n

with

|a_n|

and test whether the resulting non-negative series converges. You can use any standard convergence test on

\sum |a_n|

, such as the ratio test, root test, comparison test, or p-series test. The ratio test and root test are especially convenient because they automatically test absolute convergence.

Absolute Convergence vs. Conditional Convergence

	Absolute Convergence	Conditional Convergence
Definition	$\sum \|a_n\|$ converges	$\sum a_n$ converges but $\sum \|a_n\|$ diverges
Strength	Stronger condition — implies ordinary convergence	Weaker — series converges only because of sign cancellation
Rearrangement	Terms can be rearranged in any order without changing the sum	Rearranging terms can change the sum to any value (Riemann rearrangement theorem)
Classic example	$\sum \frac{(-1)^n}{2^n}$ (geometric, $\|r\|<1$ )	$\sum \frac{(-1)^{n+1}}{n}$ (alternating harmonic)
Tests used	Ratio test, root test, comparison test on $\sum \|a_n\|$	Alternating Series Test after absolute convergence fails

Why It Matters

Absolute convergence appears throughout calculus and analysis, especially when working with power series and determining their intervals of convergence. It matters because absolutely convergent series behave predictably — you can rearrange, regroup, or multiply them term-by-term without worrying about changing the result. Many convergence tests, including the ratio test and root test, actually test for absolute convergence, so understanding this concept is essential for series problems on AP Calculus BC exams and in college-level courses.

Common Mistakes

Mistake: Assuming that if

\sum |a_n|

diverges, then

\sum a_n

also diverges.

Correction: A divergent absolute value series only tells you the series is not absolutely convergent. The original series might still converge conditionally. Always check for conditional convergence (e.g., with the Alternating Series Test) before concluding that the series diverges.

Mistake: Confusing the sum of the original series with the sum of the absolute value series.

Correction: Even when a series converges absolutely,

\sum a_n

and

\sum |a_n|

generally have different sums. For example,

\sum \frac{(-1)^n}{2^n} = \frac{2}{3}

while

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

. Absolute convergence is about whether the absolute value series converges, not about what it converges to.

Related Terms

Series — The general concept of summing terms
Converge — What it means for a series to have a finite sum
Convergent Series — A series whose partial sums approach a limit
Convergence Tests — Methods used to determine absolute convergence
Absolute Value — The operation applied to each term
Infinite Geometric Series — A common series type that converges absolutely when |r| < 1
Ratio — The ratio test directly tests absolute convergence

Definition:	The series \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent if the series \(\sum\limits_{n = 1}^\infty {\left\| {{a_n}} \right\|} \) converges.

Example:	Determine if \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots \) is absolutely convergent.

Solution:	To find out, consider the series \[\sum\limits_{n = 1}^\infty {\left\| {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \right\|} = \sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]
	This is an infinite geometric series with ratio \(r = \frac{1}{2}\), so we know that it converges as shown. \[\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{{1 - \frac{1}{2}}} = 2\] As a result, we can conclude that \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \) converges absolutely.

Note:	The series \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots \) and \(\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \) converge to different sums. In fact, \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \) is an infinite geometric series with ratio \(r = - \frac{1}{2}\), so \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots = \frac{1}{{1 - \left( { - \frac{1}{2}} \right)}} = \frac{2}{3}\]