Intervals of Absolute and Conditional Convergence of a Series

# Intervals of Absolute and Conditional Convergence of a Series

Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. Furthermore, a convergent series $\sum_{n=1}^{\infty} a_n$ is said to be conditionally convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ diverges. We will now look at some examples of determining when a series converges absolutely, conditionally, and diverges.

## Example 1

Determine for which values of $x$ the series $\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n^24^n}$ converges absolutely, converges conditionally, and diverges.

First let's apply the Ratio test on the absolute value of the general term as follows:

(1)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{(x - 2)^{n+1}}{(n+1)^24^{n+1}} \cdot \frac{n^24^n}{(x - 2)^n} \biggr \rvert = \lim_{n \to \infty} \frac{n}{4(n+1)^2} \rvert x - 2 \rvert = \lim_{n \to \infty} \frac{1}{4} \rvert x - 2 \rvert \end{align}

We know by the ratio test that if $\frac{1}{4} \rvert x - 2 \rvert < 1$ then the series converges absolutely, that is for $-2 < x < 6$. We must now test the endpoints of this interval.

If we let $x = -2$ then $\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n^24^n} = \sum_{n=1}^{\infty} \frac{(-4)^n}{n^24^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ which converges absolutely.

If we let $x = 6$ then $\sum_{n=1}^{\infty} \frac{(4)^n}{n^24^n} = \sum_{n=1}^{\infty} \frac{1}{n^2}$ which also converges absolutely.

Therefore $\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n^24^n}$ converges absolutely for $-2 ≤ x ≤ 6$, is conditionally convergent nowhere, and divergent for all other $x$.

## Example 2

Determine for which values of $x$ the series $\sum_{n=1}^{\infty} \frac{x^n}{2^n \ln n}$ converges absolutely, converges conditionally, and diverges.

Applying the ratio test we have that:

(2)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{x^{n+1}}{2^{n+1} \ln (n+1)} \cdot \frac{2^n \ln n}{x^n} \biggr \rvert = \lim_{n \to \infty} \frac{\ln (n)}{2 \ln(n+1)} \rvert x \rvert = \frac{\rvert x \rvert}{2} \end{align}

Therefore if $\frac{\rvert x \rvert}{2} < 1$, or in other words, $-2 < x < 2$ then this series is absolutely convergent. Now let's test the endpoints of this interval.

If $x = -2$ then $\sum_{n=1}^{\infty} \frac{x^n}{2^n \ln n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln n}$ which converges conditionally.

If $x = 2$ then $\sum_{n=1}^{\infty} \frac{x^n}{2^n \ln n} = \sum_{n=1}^{\infty} \frac{1}{\ln n}$ which diverges (with $\frac{1}{\ln n } > \frac{1}{n}$ and by the comparison test).

Therefore if $-2 < x < 2$ then the series is absolutely convergent. For $x = -2$ this series is conditionally convergent, and for all other $x$ this series is divergent.

## Example 3

Determine for which values of $x$ the series $\sum_{n=1}^{\infty} \frac{(-1)^n(x - 1)^n}{2n+3}$ converges absolutely, converges conditionally, and diverges.

Applying the ratio test we have that:

(3)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{(-1)^{n+1}(x - 1)^{n+1}}{2n+5} \cdot \frac{2n+3}{(-1)^n(x-1)^n} \biggr \rvert = \lim_{n \to \infty} \frac{2n+3}{2n+5} \rvert x - 1 \rvert = \lim_{n \to \infty} \rvert x - 1 \rvert \end{align}

We note that $\rvert x - 1 \rvert < 1$ if $0 < x < 2$. Now let's test the endpoints of this interval.

If $x = 0$ then $\sum_{n=1}^{\infty} \frac{(-1)^n(- 1)^n}{2n+3} = \sum_{n=1}^{\infty} \frac{(1}{2n+3}$ diverges.

If $x = 2$ then $\sum_{n=1}^{\infty} \frac{(-1)^n(1)^n}{2n+3} = \sum_{n=1}^{\infty} \frac{(-1)^n}{2n+3}$ is conditionally convergent.

Therefore the series $\sum_{n=1}^{\infty} \frac{(-1)^n(x - 1)^n}{2n+3}$ is absolutely convergent for $0 < x < 2$, is conditionally convergent for $x = 2$, and divergent for all other $x$.