The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
 A: 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 $\tan ^{-1}x$.
A: Here is another take on this.  It may not be the most simple, but I think some may find it interesting.
Let $\chi(n)$ be the unique nontrivial Dirichlet character modulo $4$, so that by using $\frac{1}{(4n+1)(4n+3)}=\frac{1}{2}\left(\frac{1}{4n+1}-\frac{1}{4n+3}\right)$ your series equals $$\frac{1}{2}\sum_{n=1}^\infty \frac{\chi(n)}{n}=\frac{1}{2}L(1,\chi).$$  Noting that $\chi(n)=\left(\frac{d}{n}\right)$ with $d=-4$, and that the quadratic form $x^2+y^2$ is the only class with discriminant $D=-4$, we see that by the class number formula $$L(1,\chi)=\frac{2\pi}{\omega(d)\sqrt{|d|}}=\frac{\pi}{4},$$ where $\omega(d)$ is the number of symmetries of the corresponding complex lattice.  (In our case $\omega(d)=4$, because it is the Gaussian integers).
Thus the original series evaluates to $\frac{\pi}{8}$.
A: We let 
$$f(z)=\dfrac{1}{(4z+1)(4z+3)}$$
There are two poles of $f(z)$.  They are $4z_0+3=0 \implies z_0=-\frac{3}{4}$ and $4z_1+1=0 \implies z_1=-\frac{1}{4}$.
Residue calculus tells us that
$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(\text{sum of residues of }\pi\cot(\pi z)f(z))$$
We calculate the residues of the poles ($b_0$ and $b_1$ for $z_0$ and $z_1$ respectively):
$$b_0=\operatorname {Res}_{z=z_0}=\lim_{z \to z_0}\frac{(z-z_0)\pi\cot (\pi z)}{(4z+1)(4z+3)}$$
Using L'Hopital's rule we have
$$b_0=\lim_{z \to z_0} \frac{\pi\cot (\pi z)-(z-z_0)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-3\pi/4)}{4(4)}=-\frac{\pi}{8}$$
Similarly, we have 
$$b_1=\lim_{z \to z_1} \frac{\pi\cot (\pi z)-(z-z_1)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-\pi/4)}{4(4)}=-\frac{\pi}{8}$$
So
$$
\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(-\frac{\pi}{8}-\frac{\pi}{8})=\frac{\pi}{4}\implies 
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=\frac{1}{2}\frac{\pi}{4}=\frac{\pi}{8}
$$
QED
A: Once you splitted the initial fraction within the sum as below:
$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right) $$
then you may consider the following formula that is very helpful:
If $a+b+c+d=0$, 
$$\sum_{k=0}^\infty \left(\dfrac{a}{4k+1} + \dfrac{b}{4k+2}+\dfrac{c}{4k+3}+\dfrac{d}{4k+4}\right) = \dfrac{a-c}{8} \pi + \dfrac{a+c-2d}{4} \ln(2) $$
Replace the specific values you have and you're done. The limit is $\frac{\pi}{8}$.
The proof is complete.
A: $\frac{ \pi }{8}$ is what i get... 
A: Here is another method.
We have that
$$\sum_{n=0}^{\infty} x^{4n} = \dfrac1{1-x^4}$$
Integrate the above from $x=0$ to $x=t < 1$, to get $$\sum_{n=0}^{\infty} \dfrac{t^{4n+1}}{4n+1} = \int_0^{t} \dfrac{dx}{1-x^4} = \dfrac12 \left( \int_0^{t} \dfrac{dx}{1+x^2} + \int_0^{t} \dfrac{dx}{1-x^2} \right)\\ = \dfrac12 \left( \arctan(t)  + \dfrac12 \left( \int_0^t \dfrac{dx}{1+x} + \int_0^t \dfrac{dx}{1-x} \right)\right)\\
=\dfrac12 \left( \arctan(t) + \dfrac12 \left( \log(1+t) - \log(1-t)\right)\right)$$
Now multiply throughout by $t$ to get,
$$\sum_{n=0}^{\infty} \dfrac{t^{4n+2}}{4n+1} =\dfrac12 \left( t\arctan(t) + \dfrac12 \left( t\log(1+t) - t\log(1-t)\right)\right)$$
Now integrate the above from $t=0$ to $1$. Note that
$$\int_0^1 t\arctan(t) dt = \dfrac{\pi-2}{4}$$
$$\int_0^1 t\log(1+t) dt = \dfrac14$$
$$\int_0^1 t\log(1+t) dt = -\dfrac34$$
The above integrals can be evaluated with relative ease by integration by parts.
Hence, we now get that $$\sum_{n=0}^{\infty} \dfrac1{(4n+1)(4n+3)} = \dfrac12 \left( \dfrac{\pi-2}{4} + \dfrac12 \left( \dfrac14 - \left( - \dfrac34\right)\right)\right) = \dfrac{\pi}8$$
