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

$$\sum_{n\ge 0}\left(\frac1{4n+1}-\frac1{4n+3}\right)=1-\frac13+\frac15-\frac17+... =\frac\pi4$$
Precisely, consider the partial sums $S_k:=\displaystyle\sum_{n=0}^k\frac1{(4n+1)(4n+3)}$, then it is the same as the $2k^{\text{th}}$ partial sum of the Leibniz series $\displaystyle\sum_{n=0}^{2k}\frac{(-1)^n}{2n+1}$, which converges to $\pi/4$. So, your solution is $\pi/8$.
