<s>[INST] Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$ Here is a problem I need to solve:
$$ \sum_{k=2}^n  \frac{1}{k^2-1} $$
It came with another one alreay done in the same task:
$$ \sum_{k=1}^n  \frac{1}{k(k+1)} = \sum_{k=1}^n  \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n  \frac{1}{k} - \sum_{k=1}^n  \frac{1}{k+1} = \sum_{k=1}^n  \frac{1}{k} - \sum_{k=2}^{n+1}  \frac{1}{k} = 1 - \frac{1}{n+1} $$
So, the problem have to have similar solution. I can't find a proper change for the numerator. I think the numeratore could be find using factorised dominator:
$$ \sum_{k=2}^n  \frac{1}{(k-1)(k+1)} $$
 [/INST] Hint
$$\frac{1}{k^2-1}=\frac 1 2\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac 1 2\left(\frac{1}{k-1}-\frac 1 k+\frac 1 k-\frac{1}{k+1}\right)$$
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