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.

Since OP is apparently aware this is a telescoping series, here's the computation.
$$
2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)} = 
\sum_{n=2}^{\infty}\frac{2}{(n-1)(n+1)}=
\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) =\\
\left[ 
\left(1 - \frac 13\right) + \left(\frac 12 - \frac 14\right) 
+ \left(\frac 13 - \frac 15\right) + \cdots
\right] = \\
\left[ 
1 + \frac 12 + \left(\frac 13 - \frac 13\right) + \left(\frac 14 - \frac 14\right) + \cdots
\right] = \frac 32
$$
