Absolute and Conditional Convergence

# Absolute and Conditional Convergence

Sometimes we want to decide whether a series is convergent or divergent, but the sequence isn't necessarily positive. We will learn a technique to evaluate series of this nature but we must first look at a very important definition regarding convergence first.

 Definition: Let $\sum_{n=1}^{\infty} a_n$ is a convergent series. We say that this series is Absolutely Convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. We say the original series is Conditionally Convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is not convergent.

We will now look directly into an important theorem that may seem obvious.

 Theorem 1: If the series $\sum_{n=1}^{\infty} \mid a_n \mid$ converges then the series $\sum_{n=1}^{\infty} a_n$ also converges.
• Proof of Theorem: Suppose that $\sum_{n=1}^{\infty} \lvert a_n \rvert$ is converges. Let the sequence $\{ b_n \}$ be defined such that for all $n \in \mathbb{N}$, $b_n = a_n + \mid a_n \mid$. Now we note that $-\mid a_n \mid ≤ a_n ≤ \mid a_n \mid$ and so $0 ≤ a_n + \mid a_n \mid ≤ 2\mid a_n \mid$ so then $0 ≤ b_n ≤ 2\mid a_n \mid$. Therefore $\sum_{n=1}^{\infty} b_n$ converges by the comparison test, and so it follows that $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} b_n - \sum_{n=1}^{\infty} \mid a_n \mid$ must also converge. $\blacksquare$

From theorem 1, we see that if a series is absolutely convergent then it is convergent, which gives us a test for convergence that does NOT require the series to be positive. Of course, if a series is already positive, then testing for absolute convergence is the same thing as testing for regular convergence.

## Example 1

Determine if the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is absolutely convergent or conditionally convergent.

We know by the p-series test that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is in fact convergent since $p = 2 > 1$. Now notice that this series is positive for $n ≥ 1$, $n \in \mathbb{N}$, and thus it follows that:

(1)
\begin{align} \sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty} \biggr \rvert \frac{1}{n^2} \biggr \rvert \end{align}

So then the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is absolutely convergent.

## Example 2

Determine if the series $\sum_{n=1}^{\infty} \frac{\sin \left (n\frac{\pi}{2} \right )}{n}$ is absolutely convergent or conditionally convergent.

We note that if $n$ is odd then $\sin \left (n\frac{\pi}{2} \right ) = 1$ and if $n$ is even then $\sin \left (n\frac{\pi}{2} \right ) = -1$. Thus we will replace $\sin \left (n\frac{\pi}{2} \right )$ with $(-1)^n$, that is $\sum_{n=1}^{\infty} \frac{\sin \left (n\frac{\pi}{2} \right )}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

Now we note that $\sum_{n=1}^{\infty} \biggr \rvert \frac{(-1)^n}{n} \biggr \rvert = \sum_{n=1}^{\infty} \frac{1}{n}$ which we deduce from the p-series test to divergent. So the original series is not absolutely convergent. We don't know whether $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ is convergent or divergent though. This series does converge, and we will be able to solve this later.