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$$\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$$.
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.
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No, the bottom identity will NOT help at all. – Christopher K May 09 '14 at 12:57
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1Note $4n^4+1$ factors, then maybe you can use partial fractions. – Gerry Myerson May 09 '14 at 13:02
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This might help: $\frac{1}{4n^4+1}=\frac{1}{(2n^2+1)^2-4n^2}=\frac{1}{(2n^2+2n+1)(2n^2-2n+1)}$ – barak manos May 09 '14 at 13:27
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1Duplicate of Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ – Grigory M May 09 '14 at 13:36
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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).
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