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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.

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    $\begingroup$ See if you can utilize the identity $\pi\coth(\pi x) = \frac1{x} + 2\sum_{n=1}^\infty \frac{x}{x^2 + n^2}$. $\endgroup$
    – user43208
    Commented May 18 at 2:27
  • $\begingroup$ @user43208 Thanks, one way I can think of to use the identity you just gave, for $x=1$ we have $\sum_{n=1}^{\infty} \frac{2}{n^2 + 1} = \sum_{n=1}^{\infty} \frac{8}{n^4 + 4} = \pi \coth(\pi)-1$ (since we know the answer beforehand). But unfortunately, I don't know how to prove this two sums are equal. $\endgroup$
    – Briston
    Commented May 18 at 16:25

2 Answers 2

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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 $\pi \coth \pi - 1$.

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Too long for a comment

$$S_2=\sum_{n=1}^\infty\frac8{n^4+4}=-1+\sum_{n=-\infty}^\infty\frac4{n^4+4}=-1+4S(a=\sqrt2)$$ where we denote $$S(a)=\sum_{n=-\infty}^\infty\frac1{n^4+a^4}$$ A standart way of evaluation such sums is the integration in the complex plane along a big circle of the radius $R\to\infty$ (contour $C$) $$\oint_C\frac{\pi\cot\pi z}{z^4+a^4}dz=2\pi i\sum\operatorname{Res}\left(\frac{\pi\cot\pi z}{z^4+a^4}\right)\to0\,\,\text{at}\,\,R\to\infty$$ Simple poles at $z=0;\pm1;\pm2...$ give us $S(a)$; we also have poles in the points $z=e^{\pm\frac{\pi i}4}a;\,e^{\pm\frac{3\pi i}4}a$

Therefore, $$S(a)+\underset{z=e^{\pm\frac{\pi i}4}a;\,e^{\pm\frac{3\pi i}4}a}{\operatorname{Res}}\left(\frac{\pi\cot\pi z}{z^4+a^4}\right)=0$$ In our case $a=\sqrt2$, what gives considerable simplification. Using $\,\cot\big(\pi(\pm1+i)\big)=\cot (\pi i)=-i\coth\pi$ $$S(\sqrt2)+\underset{z=1\pm i;\,-1\pm i}{\operatorname{Res}}\left(\frac{\pi\cot\pi z}{(z-(1+i))(z-(1-i))(z-(-1+i))(z-(-1-i))}\right)=0$$ $$S(\sqrt2)=\frac\pi 8\coth\pi\cdot2\cdot\frac{1-i+1-i}{1-i^2}=\frac\pi4\coth\pi$$ $$\boxed{\,\,S_2=-1+\pi\coth\pi\,\,}$$

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