\sum_{n \geq 2}{\frac{1}{n^2-1}}=-\frac{1}{2} \sum_{n \geq 2}{ \frac{1}{n+1}} + \frac{1}{2} \sum_{n \geq 2}{\frac{1}{n-1}} 
= -\frac{1}{2} \sum_{n \geq 2}{ \frac{1}{n+1}} + \frac{1}{2} ( \sum_{m \geq 2}{\frac{1}{m+1}+1+\frac{1}{2}}) =
 -\frac{1}{2} \sum_{n \geq 2}{ \frac{1}{n+1}} + \frac{1}{2} \sum_{m \geq 2}{\frac{1}{m+1}}+\frac{1}{2}(1+\frac{1}{2}) 
= \frac{3}{4}