Absolute Convergence – Definition, Condition, and Examples

A series’ absolute convergence showcases a stronger type of convergence. Understanding this topic will add one more tool kit to our convergence tests.

The absolute convergence test also helps us to check an alternating series test’s convergence when the divergence or alternating series test is not helpful. Testing a series for absolute convergence will depend on the result of adding the absolute values of each of the terms.

In this article, we’ll briefly discuss what makes absolute convergence special. We’ll also learn how to check a series for absolute convergence. By the end of our discussion, we hope that you can answer the sample problems we’ve prepared for you as well!

What is absolute convergence?

The absolute convergence has a higher degree of convergence than the regular convergence. Given a series, $\sum_{n = 0}^{\infty} a_n$, when we take the absolute value of each of the terms, we’ll have the series shown below.

\begin{aligned}\sum_{n = 0}^{\infty} |a_n| &= |a_1| + |a_2| + |a_3| + …\end{aligned}

When $\boldsymbol{\sum_{n = 0}^{\infty} |a_n|}$ is convergent, the series, $\boldsymbol{\sum_{n = 0}^{\infty} a_n}$ will exhibit absolute convergence.

Let’s say we have a series with terms that contain both positive and negative values such as the series shown below.

\begin{aligned}3 – \dfrac{3}{2} + \dfrac{3}{4} – \dfrac{3}{8} + … &= \sum_{n = 0}^{\infty} 3\left(-\dfrac{1}{2}\right)^n \end{aligned}

This series is a geometric series with a common ratio of $\dfrac{1}{2}$, so the series is convergent.  Now, what happens if the series has no alternating signs?

\begin{aligned}3 + \dfrac{3}{2} + \dfrac{3}{4}+ \dfrac{3}{8} +… &= \sum_{n = 0}^{\infty} \left|3\left(-\dfrac{1}{2}\right)^n \right|\\&= \sum_{n = 0}^{\infty} 3\left(\dfrac{1}{2}\right)^n\end{aligned}

Since the common ratio is still less than $1$, this geometric series will still be convergent. The series shown above will be absolutely convergent as well.

Absolute convergence test

Let’s summarize the process of testing the series for absolute convergence. Suppose that we have a series, $\sum_{n = 0}^{\infty} a_n$, we can use the absolute values of its terms to determine its absolute convergence.

When $\sum_{n = 0}^{\infty} |a_n|$ is divergent, the series may be convergent. In addition, when we can show that $\sum_{n = 0}^{\infty} a_n$ is absolutely convergent, the series is of course, convergent as well.  We left the proof of this statement in the exercises for you as well!

How to test for absolute convergence?

When testing a series for absolute convergence, we can still use the different converges tests we’ve learned in the past. Take the absolute values of each of the terms of the series and see if the resulting series is convergent.

The two most common convergence tests used to check a series for absolute convergence are the ratio test and the root test. Here’s a quick recap of these two convergence tests:

Once we’ve shown that $\sum_{n = 0}^{\infty} |a_n|$ is convergent, we can conclude that the series is indeed absolutely convergent. We’ve prepared some exercises for you to work on when you’re ready!

Example 1

Prove the statement: “ If $\sum_{n = 0}^{\infty} a_n$ is absolutely convergent, then the series is also convergent.”.

Solution

Recall that $|a_n|$ will either be $a_n$ or $-a_n$ based on the definition of absolute values. Hence, it is safe for us to conclude that

\begin{aligned}0 \leq a_n + |a_n| \leq 2|a_n|.\end{aligned}

Since $|a_n|$ is absolutely convergent, 2|a_n| will also be absolutely convergent. Using the comparison test, one of the convergence tests we’ve learned in the past, we can conclude that $\sum_{n = 0}^{\infty} (a_n + |a_n|)$ is also convergent. Hence, we have

\begin{aligned} \sum_{n = 0}^{\infty} a_n &= \sum_{n = 0}^{\infty} (a_n + |a_n|) -\sum_{n = 0}^{\infty} |a_n| \\\end{aligned}

and given that the two series are convergent, their difference will also be convergent. This shows that when $\sum_{n = 0}^{\infty} a_n$ is absolute convergent, $\sum_{n = 0}^{\infty} a_n$ is also convergent.

Example 2

Confirm that the series, $\sum_{n =1}^{\infty} \dfrac{n!}{n^{n}}$, is absolutely convergent. Use the fact that $\lim_{n \rightarrow \infty} \left(\dfrac{n}{n + k}\right)^n = e^{-n}$.

Solution

Since the series has $n$ in the bases of both the numerator and denominator, let’s use the ratio test to check the series for absolute convergence. Take the ratio of the $n$th and $(n + 1)$th terms then evaluate the limit of the resulting expression as $n \rightarrow \infty$.

\begin{aligned}\lim_{n \rightarrow \infty}\left|\dfrac{a_{n +1}}{a_n} \right| &= \lim_{n \rightarrow \infty}\left|\dfrac{\dfrac{(n +1)!}{(n+1)^{(n +1)}}}{\dfrac{n!}{n^n}} \right|\\&= \lim_{n \rightarrow \infty}\left|\dfrac{(n +1)!}{(n+1)^{(n +1)}} \cdot \dfrac{n^n}{n!}\right|\\&=\lim_{n \rightarrow \infty}\left|\dfrac{(n +1)n!}{(n+1)^{(n +1)}} \cdot \dfrac{n^n}{n!}\right|\\&=\lim_{n \rightarrow \infty}\left|\dfrac{\cancel{(n +1)}\cancel{n!}}{(n+1)^{n} \cdot \cancel{(n + 1)}} \cdot \dfrac{n^n}{\cancel{n!}}\right|\\ &= \lim_{n \rightarrow \infty} \left| \dfrac{n^n}{(n + 1)^n}\right| \\&= \lim_{n \rightarrow \infty} \left| \left(\dfrac{n}{n + 1}\right)^n\right| \end{aligned}

Simplify the limit using the common limit, $\lim_{n \rightarrow \infty} \left(\dfrac{n}{n + k}\right)^n = e^{-k}$, where $k = 1$.

\begin{aligned}\lim_{n \rightarrow \infty} \left| \left(\dfrac{n}{n + 1}\right)^n\right| &= e^{-1}\\&= \dfrac{1}{e}\\&\approx 0.36787 \end{aligned}

Since the resulting limit is less than $1$, through the ratio test, we can conclude that the series, $\sum_{n = 1}^{\infty} \dfrac{n!}{n^{n}}$, is absolutely convergent.

Example 3

Confirm that the series, $\sum_{n =1}^{\infty} \dfrac{n^{-3(n -1/3)}}{4^{2n}}$, is absolutely convergent.

Solution

Since both the numerator and denominator have $n$ in their exponents, so the best test to use is the root test. Take the $n$th root of $|a_n| = \left|\dfrac{n^{-3(n -1/3)}}{4^{2n}}\right|$ then evaluate the limit of $\sqrt[n]{|a_n|}$ as $n \rightarrow \infty$.

\begin{aligned}\lim_{n \rightarrow \infty} \sqrt[n]{\left|\dfrac{n^{-3(n -1/3)}}{4^{2n}}\right|} &= \lim_{n \rightarrow \infty} \sqrt[n]{\left|\dfrac{n^{-3n + 1}}{(4^{2})^{n}}\right|}\\ &= \lim_{n \rightarrow \infty} \sqrt[n]{\left|\dfrac{n^{-3n} \cdot n}{(4^{2})^n}\right|}\\&=\lim_{n \rightarrow \infty} \dfrac{\sqrt[n]{|n|}\sqrt[n]{|n^{-3}|^n}}{\sqrt[n]{|(4^2)^n|}}\\&= \lim_{n \rightarrow \infty} \dfrac{\sqrt[n]{|n|} n^{-3}}{4^2}\end{aligned}

We’ve learned in the past that $\lim_{n \rightarrow \infty} \sqrt[n]{n} = 1$. Use this to simplify the limit of $\dfrac{\sqrt[n]{|n|} n^{-3}}{4^2}$ as $n$ approaches infinity.

\begin{aligned}\lim_{n \rightarrow \infty} \dfrac{\sqrt[n]{|n|} n^{-3}}{4^2} &= \lim_{n \rightarrow \infty} \dfrac{\sqrt[n]{|n|} \dfrac{1}{n^3}}{4^2}\\&= \dfrac{1 \cdot 0}{16}\\&= 0\end{aligned}

Since the limit of the expression is less than $1$, through the root test, we can conclude that the series,  $\sum_{n =1}^{\infty} \dfrac{n^{-3(n -1/3)}}{4^{2n}}$, is absolutely convergent.

Practice Questions

1. Determine whether the series, $\sum_{n =1}^{\infty} \dfrac{n! }{80^n}$, is convergent, absolutely convergent, or divergent.
2. Determine whether the series, $\sum_{n =1}^{\infty} \dfrac{\sin (3n) }{3^n}$, is convergent, absolutely convergent, or divergent.
3. Determine whether the series, $\sum_{n =1}^{\infty} \dfrac{(-3)^n}{n^n}$, is convergent, absolutely convergent, or divergent.
4. Determine whether the series, $\sum_{n =1}^{\infty} \left(\dfrac{-2n}{n + 1}\right)^{5n}$, is convergent, absolutely convergent, or divergent.
5. Determine whether the series, $\sum_{n =1}^{\infty} (-1)^n\sin\left(\dfrac{\pi}{n}\right)$, is convergent, absolutely convergent, or divergent.

Answer Key

1. $\sum_{n =1}^{\infty} \dfrac{n! }{80^n}$ is divergent.
2. $\sum_{n =1}^{\infty} \dfrac{\sin (3n) }{3^n}$ is absolutely convergent.
3. $\sum_{n =1}^{\infty} \dfrac{(-3)^n}{n^n}$ is absolutely convergent.
4. $\sum_{n =1}^{\infty} \left(\dfrac{-2n}{n + 1}\right)^{5n}$ is divergent.
5. $\sum_{n =1}^{\infty} (-1)^n\sin\left(\dfrac{\pi}{n}\right)$ is convergent.