Sum of infinite telescoping series $\sum_{r=2}^\infty \frac{1}{r^2-1}$?

Question

How do I find sum of $\sum_{r=2}^\infty \frac{1}{r^2-1}$? The answer given in my book is 3/4. I can decompose the general term into $(\frac{1}{r-1}-\frac{1}{r+1})$ multiplied by 1/2 but since it is infinite I don't know what to do next. How should I do this?

Answer 1

Let consider

$$\sum_{r=2}^n \frac{2}{r^2-1}=\sum_{r=2}^n(\frac{1}{r-1}-\frac{1}{r+1})=\sum_{r=2}^n \frac{1}{r-1}-\sum_{r=2}^n \frac{1}{r+1}=$$$$=\sum_{r=1}^{n-1} \frac{1}{r}-\sum_{r=3}^{n+1} \frac{1}{r}=1+\frac12-\frac1n-\frac1{n+1}$$

and then take the limit.

Answer 2

Hint: to calculate the partial sums add the first, third, fifth,... terms and the second, fourth,.. terms. The answer is $\frac 1 2(1+\frac 1 2)=\frac 3 4$.