Show that $\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$ Show that 
$$\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$$.
I am thinking of using Fourier series and Parseval's identity to tackle this, I tried $x^4$, $4x^4+1$, however these did not work quite well.

It can be shown using Poisson summation formula that 
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b^2}=\frac{\pi}{b}\tanh \pi b.$$
Setting $b_{\pm}=e^{\pm i\pi/4}c$ in this formula, we get
\begin{align}
\sum_{n\in\mathbb{Z}}\frac{1}{n^4+c^4}&=\frac{1}{b_+^2-b_-^2}\left(\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b_-^2}-\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b_+^2}\right)=\\ &=\frac{\pi}{2ic^2}\left[\frac{\tanh \pi b_-}{b_-}-\frac{\tanh \pi b_+}{b_+}\right].\tag{1}
\end{align}
On the other hand
$$\tanh\frac{\pi e^{\pm i\pi/4}}{\sqrt{2}}=\tanh\frac{\pi(1\pm i)}{2}=\coth\frac{\pi}{2}.$$
Hence, setting $c=1/\sqrt2$ in (1), we get
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^4+\frac14}=2\pi\coth\frac{\pi}{2}.$$
From this one easily finds
$$\boxed{\displaystyle\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac14\sum_{n=0}^{\infty}\frac{1}{n^4+\frac14}=\frac{\pi}{4}\coth\frac{\pi}{2}+\frac12}$$
(Note the misprint $\coth\rightarrow\tanh$ in your question).
