Answer
Convergent
$S=\frac{3}{4}$
Work Step by Step
Given
$$ \sum_{k=2}^{\infty} \frac{1}{k^{2}-1}$$
Since by using partial fractions
\begin{aligned}
\sum_{k=2}^{\infty} \frac{1}{k^{2}-1} &=\sum_{k=2}^{\infty} \frac{1}{(k-1)(k+1)} \\
&=\sum_{k=2}^{\infty} \frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right) \\ &=\lim _{n \rightarrow \infty} \sum_{k=2}^{n} \frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right) \\
& =\lim_{n\to\infty} \frac{1}{2}\bigg\{ \left(\frac{1}{2-1}-\frac{1}{2+1}\right)+\left(\frac{1}{3-1}-\frac{1}{3+1}\right)\ \ \ \ + \left(\frac{1}{4-1}-\frac{1}{4+1}\right)+ \cdots+\left( \frac{1}{n-1}-\frac{1}{n+1}\right)\bigg\} \\
&=\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-0-0\right) \\
&=\frac{3}{4}
\end{aligned}
The series is convergent
