Show that this equation is equal to $\sum_{k=2}^{2n+1}\frac{2}{k^2-1}$

Question

Show that this... $\sum_{k=2}^{2n+1}\frac{2}{k^2-1}= \frac{3}{2}-\frac{1}{2n+1}-\frac{1}{2n+2}$

I have already arrived at... 1+ $\frac{1}{2}$ + ($\frac{1}{2n-1}$ - $\frac{1}{2n+1}$) + ($\frac{1}{2n}$ - $\frac{1}{2n+2}$)

But I am unaware as to how to get the last part of the equation... $\frac{1}{2n+1}-\frac{1}{2n+2}$

Answer 1

HINT:

Note that

$$\frac2{k^2-1}=\frac{1}{k-1}-\frac1{k+1}=\left(\frac{1}{k-1}-\frac1k\right)+\left(\frac1k-\frac{1}{k+1}\right)$$

Now telescope two summations.