Question

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.

$\sum _{k=2}^{\infty }\frac{{\left(-1\right)}^k\ln k}{k^2-1}$

Short answer

The series converges absolutely.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{k=2}^{\infty }\frac{{\left(-1\right)}^k\ln k}{k^2-1}$

Step 2. Use the alternating series test.

According to the alternating series test, assume $\left\{a_k\right\}$be the sequence of positive numbers.

If $a_{k+1}<a_k$for every $k\ge 1$and $\underset{k\to \infty }{\lim }{a}_k=0$.

Then the alternating series $\sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}{a}_k$and $\sum _{k=1}^{\infty }{\left(-1\right)}^k{a}_k$both converge.

Consider the series $\sum _{k=1}^{\infty }{\left(-1\right)}^k{a}_k,\sum _{k=2}^{\infty }\frac{{\left(-1\right)}^k\ln k}{k^2-1}$.

Clearly, we can see that the series is decreasing as $a_{k+1}<a_k$.

Step 3. Find the value of the limit.

The value of$\underset{k\to \infty }{\lim }{a}_k$,

localid="1649133648893" $$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{\ln k}{k^2-1}\right) \\ \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{{\displaystyle \frac{1}{k}}}{2k}\right) \\ \underset{k\to \infty }{\lim }{a}_k=0 \end{gathered}$$

As, all three conditions are satisfied. Therefore, by alternating series test, the test localid="1649133903784" $\sum _{k=2}^{\infty }\frac{{\left(-1\right)}^k\ln k}{k^2-1}$is convergent.

Step 4. Rewrite the series.

The series $\sum _{k=2}^{\infty }\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|$can be written as,

$$\begin{gathered} \sum _{k=2}^{\infty }\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|=\sum _{k=2}^{\infty }\left|\frac{\ln k}{k^2-1}\right| \\ \sum _{k=2}^{\infty }\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|=\sum _{k=2}^{\infty }\frac{\ln k}{k^2-1} \end{gathered}$$

The terms of the series $\sum _{k=2}^{\infty }\frac{\ln k}{k^2-1}$are positive.

The series $\sum _{k=1}^{\infty }{b}_k$for the series $\sum _{k=2}^{\infty }\frac{\ln k}{k^2-1}$is given by,

$\sum _{k=2}^{\infty }\frac{\ln k}{k^2-1}\le \sum _{k=1}^{\infty }\frac{\ln k}{k^2}$

Step 5. Find the value of the limit.

The value of $\underset{k\to \infty }{\lim }\frac{\ln k}{k^2}$,

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{\ln k}{k^2}=\underset{k\to \infty }{\lim }\left(\frac{{\displaystyle \frac{1}{k}}}{2k}\right) \\ =0 \end{gathered}$$

The series role="math" localid="1649136943547" $\underset{k=1}{\lim }\frac{\ln k}{k^2}$is convergent. Therefore, role="math" localid="1649136949081" $\underset{k=2}{\lim }\frac{\ln k}{k^2-1}$is convergent.

Hence, $\underset{k=2}{\lim }\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|$is convergent.

Thus, the series$\sum _{k=2}^{\infty }\frac{{\left(-1\right)}^k\ln k}{k^2-1}$converges absolutely.