<s>[INST] find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$ 
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$

I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
 [/INST] Notice that the first few terms are:
$(1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)\cdots$
which gives: $1/2+1/4=3/4$. 
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