How is $2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$ calculated? $$2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$$
I cant figure out why this is $\frac64$. I try to use telescopic series without success.

Ok well first of all $\frac{6}{4}=\frac{3}{2}$.
The series telescopes as has been pointed out, using that formula we have 
$$\left(\frac{1}{1}-\frac{1}{3}\right)+
\left(\frac{1}{2}-\frac{1}{4}\right)+
\left(\frac{1}{3}-\frac{1}{5}\right)+
\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots =1+\frac{1}{2}=\frac{3}{2}$$
