Hint on computing the series $\sum_{n=2}^\infty \frac{1}{n^2-1}$.

Solution 1:

Use partial fraction decomposition to represent the sum and notice that it is a telescoping series.

Solution 2:

Since $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac12\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ as you already have, then write for $N\in \mathbb N$
\begin{align}\sum_{n=2}^{N}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)&=\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\dots+\left(\frac1{N-1}-\frac1{N+1}\right)\\[0.2cm]&=1+\frac12+\left(\frac13-\frac13\right)+\dots+\left(\frac1{N-1}-\frac1{N-1}\right)-\frac1N-\frac1{N+1}\\[0.2cm]&=\frac32-\frac1N-\frac1{N+1}\end{align} In other words, this sum telescopes. Now let $N\to \infty$ (and of course do not forget 1/2 in front of the sum) to conclude that $$\sum_{n=2}^{+\infty}\frac{1}{n^2-1}=\lim_{N\to+\infty} \frac12\sum_{n=2}^N\left(\frac{1}{n-1}-\frac1{n+1}\right)=\frac12\lim_{N\to+\infty}\left(\frac32-\frac1N-\frac1{N+1}\right)=\frac34$$

Solution 3:

hint: $\dfrac{1}{n^2-1} = \dfrac{1}{2}\cdot \left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right) = \dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right) + \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$. From this you see that there are $2$ sums you calculate, and using telescoping the first sum is $\dfrac{1}{2}$, and the second is $\dfrac{1}{4}$, thus the answer is $\dfrac{3}{4}$ as claimed.