91Ó°ÊÓ

Consider the given question,

$\underset{k=2}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k\ln k}{k^2-1}$

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 $kâ‰\yen 1$and $\underset{k→∞}{\lim }{a}_k=0$.

Then the alternating series $\underset{k=1}{\overset{∞}{∑}}{\left(-1\right)}^{k+1}{a}_k$and $\underset{k=1}{\overset{∞}{∑}}{\left(-1\right)}^k{a}_k$both converge.

Consider the series $\underset{k=1}{\overset{∞}{∑}}{\left(-1\right)}^k{a}_k,\underset{k=2}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k\ln k}{k^2-1}$.

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

The value of$\underset{k→∞}{\lim }{a}_k$,

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

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

The series $\underset{k=2}{\overset{∞}{∑}}\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|$can be written as,

$$\begin{gathered} \underset{k=2}{\overset{∞}{∑}}\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|=\underset{k=2}{\overset{∞}{∑}}\left|\frac{\ln k}{k^2-1}\right| \\ \underset{k=2}{\overset{∞}{∑}}\left|\frac{{\left(-1\right)}^k\ln k}{k^2-1}\right|=\underset{k=2}{\overset{∞}{∑}}\frac{\ln k}{k^2-1} \end{gathered}$$

The terms of the series $\underset{k=2}{\overset{∞}{∑}}\frac{\ln k}{k^2-1}$are positive.

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k$for the series $\underset{k=2}{\overset{∞}{∑}}\frac{\ln k}{k^2-1}$is given by,

$\underset{k=2}{\overset{∞}{∑}}\frac{\ln k}{k^2-1}≤\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}$

The value of $\underset{k→∞}{\lim }\frac{\ln k}{k^2}$,

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

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$\underset{k=2}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k\ln k}{k^2-1}$converges absolutely.