The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}

Hint:
$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right) $$
Then note that
$$\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}} = \frac{1}{{2\left( {2n+1} \right) + 1}} - \frac{1}{{2\left( {2n} \right) + 1}}$$
from where
$$\sum\limits_{n = 1}^m {\left( {\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}}} \right)}  = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} +  -  \cdots  + \frac{1}{{2(2m + 1) + 1}} - \frac{1}{{2(2m) + 1}} $$
then recall the series for $\tan ^{-1}x$.
