Question: 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:

$$ \sum\_{k=2}^n \frac{1}{(k-1)(k+1)} $$

Answer: 
**Hint**

$$\frac{1}{k^2-1}=\frac 1 2\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac 1 2\left(\frac{1}{k-1}-\frac 1 k+\frac 1 k-\frac{1}{k+1}\right)$$
$\frac{1}{(k-1)(k+1)}=\frac{1}{2}\frac{1}{k-1}-\frac{1}{2}\frac{1}{k+1}$