Calculate this sum $\sum \frac{1}{(4n+1)(4n+3)}$ I'm having troubles to calculate this sum: $\sum \frac{1}{(4n+1)(4n+3)}$. I'm trying to use telescopic series, without success:
$\sum \frac{1}{(4n+1)(4n+3)}=1/2\sum \frac{1}{(4n+1)}-\frac{1}{(4n+3)}$
I need help here
Thanks a lot

$$\begin{align} \frac{1}{2} \sum_{k=0}^{\infty} \left ( \frac{1}{4 k+1} - \frac{1}{4 n+3} \right ) &= \frac{1}{2} \left ( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots \right )\\ &= \frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \\ &= \frac{\pi}{8}  \end{align}$$
