# 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 $\pi/8$.
\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}
Just consider the series $$\sum_{n\geq1}\dfrac{\mathbb{i}^n}{n}=-\ln(1-\mathbb{i})=-\dfrac{\ln 2}{2}+\dfrac{\pi}{4}\mathbb{i}$$ hence,
$$\sum_{n\ge 0}\left(\frac1{4n+1}-\frac1{4n+3}\right)=\sum_{n\geq0}\dfrac{(-1)^n}{2n+1}=\Im\left(\sum_{n\geq1}\dfrac{\mathbb{i}^n}{n}\right)=\dfrac{\pi}{4}$$