Answer
Converges
Work Step by Step
Given $$\sum_{n=4}^{\infty} \frac{1 }{n^2-1}$$
Since $f(n)= \frac{1 }{n^2-1},\ \ f'(n) =-\frac{2n}{\left(n^2-1\right)^2}$, is positive, decreasing and continuous for $n\geq4$, then by using the integral test
\begin{aligned} \int_{4}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{4}^{t} \frac{1 }{n^2-1}d n \\
&=\frac{1}{2}\lim _{t \rightarrow \infty} \int_{4}^{t} \frac{1 }{n -1}-\frac{1 }{n +1}d n \\
&=\frac{1}{2}\left.\lim _{t \rightarrow \infty}\ln\frac{n-1}{n+1}\right|_{4} ^{t} \\
&= \frac{1}{2} \lim _{t \rightarrow \infty}\ln\frac{t-1}{t+1} -\ln\frac{3}{5}\\ &=\frac{-1}{2}\ln\frac{3}{5} \end{aligned}
Thus the series converges.
