Find $\sum_{n=1}^\infty (a_{2n}^2+b_{2n}^2)\ \ \ \text{where}\ \ \ e^x=a_0+\sum_{n=1}^\infty a_n\cos(nx)+b_n\sin(nx) $ Find the sum:
$$
S=\sum_{n=1}^\infty (a_{2n}^2+b_{2n}^2)\ \ \ \text{where}\ \ \ e^x=a_0+\sum_{n=1}^\infty a_n\cos(nx)+b_n\sin(nx)
$$
I found out that (here is the proof)
$$
a_n = \frac{2(-1)^n\sinh(\pi)}{\pi(1+n^2)},\ \ \ \ \ b_n = -\frac{2n(-1)^n\sinh(\pi)}{\pi(1+n^2)};
$$
It follows that
$$
a_{2n}^2+b_{2n}^2=\frac{4\sinh^2(\pi)}{\pi^2(1+4n^2)}
$$

So, what I basically have to do is to find the sum of this series:
$$
\sum_{n=1}^\infty\frac{1}{1+4n^2}
$$

And I must do that without the following identity:
$$
\sum_{k=1}^\infty\frac{1}{k^2-x^2}=\frac{1}{2x^2}-\frac{\pi\cot(\pi x)}{2x}
$$
$$$$
P.S. If I were to find $\sum_{n=1}^\infty\frac{1}{1+n^2}$, I could easily do that by applying Parseval's identity to the Fourier series of $e^x$. But unfortunately, here, things are a bit more complicated.
 A: METHODOLOGY $1$:  CONTOUR INTEGRATION
We can evaluate the series $\displaystyle \sum_{n=1}^\infty \frac1{4n^2+1}$ by analyzing the contour integral $\displaystyle \oint_{|z|=N+1/2} \frac{\cot(\pi z)}{4z^2+1}\,dz$.
The function $\displaystyle f(z)= \frac{\cot(\pi z)}{4z^2+1}$ has poles at $z=n$, for all $n\in \mathbb{Z}$ and at $z=\pm i/2$.  The residue theorem then guarantees that for $N\ge 1$
$$\begin{align}
\oint_{|z|=N+1/2} \frac{\cot(\pi z)}{4z^2+1}\,dz&=2\pi i\sum_{n=-N}^N\text{Res}\left(\frac{\cot(\pi z)}{4z^2+1},z=n \right)\\\\
&+2\pi i  \text{Res}\left(\frac{\cot(\pi z)}{4z^2+1},z=i/2 \right)\\\\
&+2\pi i  \text{Res}\left(\frac{\cot(\pi z)}{4z^2+1},z=-i/2 \right)\\\\
&=2\pi i \left(\sum_{n=-N}^N \frac{1}{\pi(4n^2+1)}\sum_{n}\right)+\frac1{2i}\cot(i\pi/2)\tag1
\end{align}$$
I showed in This Answer that $|\cot(\pi z)|$ is bounded on $|z|=N+1/2$, $N\in \mathbb{N}$.  It is then straightforward to show that as $N\to \infty$ the integral on the right-hand side of $(1)$ approaches $0$.  Therefore, we conclude
$$\sum_{n=1}^\infty \frac1{4n^2+1}=\frac\pi 4 \coth(\pi/2)-\frac12$$


METHODOLOGY $2$:  APPLICATION OF PARSEVAL'S FORMULA
We can evaluate the series of interest using Fourier series and Parseval's formula.  The Fourier series for $e^x$ on $[0,\pi]$ is
$$e^x=\frac{\alpha_0}{2}+\sum_{n=1}^\infty\left(\alpha_n\cos(2nx)+\beta_n\sin(2nx)\right)$$
where the coefficients are given by
$$\begin{align}
\alpha_n&=\frac{2(e^\pi -1)}{\pi(1+4n^2)}\\\\
\beta_n&=-\frac{4n(e^\pi -1)}{\pi(1+4n^2)}
\end{align}$$
Now apply Parseval's formula to find the value of the series $\sum_{n=1}^\infty \frac1{4n^2+1}$ and use it to obtain the coveted result.
