Evaluate
$$
\sum_{n=1}^{\infty} \frac{n+8}{n^{4}+4}
$$
According to WolframAlpha it is $\pi\coth(\pi) – \dfrac{5}{8}$
My attempt:
I tried to separate $\dfrac{n}{n^{4}+4}$ and $\dfrac{8}{n^{4}+4}$. Now the first term is $\dfrac{3}{8}$ since
$$
n^4 + 4 = \bigl((n+1)^2 + 1\bigr)\bigl((n-1)^2 + 1)\bigr).
$$
With partial fraction decomposition, this turns into a simple telescoping sum.
However I have no idea what to do with the second term. Only thing I can think of is
$$
\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}
$$
since it is similar looking, but I could not find a way to make use of it.
I'll be glad for any help. Thanks in advance.
Best Answer
Partial fraction decomposition yields \begin{align} \frac{8}{n^4+4} &= \frac{1+i}{2(n + 1 + i)} + \frac{1-i}{2(n + 1 - i)} - \frac{1-i}{2(n - 1 + i)} - \frac{1+i}{2(n - 1 - i)} \\ &= \frac{1}{2}\left(\frac{1}{n + 1 + i} + \frac{1}{n + 1 - i} - \frac{1}{n - 1 + i} - \frac{1}{n - 1 - i}\right) \\&+ \left(\frac{i}{2(n + 1 + i)} - \frac{i}{2(n + 1 - i)}\right) + \left(\frac{i}{2(n - 1 + i)} - \frac{i}{2(n - 1 - i)}\right) \\ &= \frac{1}{2}\left(\frac{1}{n + 1 + i} + \frac{1}{n + 1 - i} - \frac{1}{n - 1 + i} - \frac{1}{n - 1 - i}\right) \\&+ \frac{1}{(n+1)^2+1} + \frac{1}{(n-1)^2+1}. \end{align} The first grouping telescopes to $-1/2$, and the rest sums to \begin{align} \sum_{n=1}^\infty \left(\frac{1}{(n+1)^2+1} + \frac{1}{(n-1)^2+1}\right) &= \frac{1}{0^2+1} + \frac{1}{1^2+1} + 2\sum_{n=2}^\infty \frac{1}{n^2+1} \\ &= \frac{1}{2} + 2\sum_{n=1}^\infty \frac{1}{n^2+1}, \end{align} so $$\sum_{n=1}^\infty \frac{8}{n^4+4} = 2\sum_{n=1}^\infty \frac{1}{n^2+1},$$ which is known to be $\pi \coth \pi - 1$.