Evaluation of $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration

Question

I don't succed to evaluate this sum $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration because using fraction method is too hard for me and i have got that is equal $\frac 3 8$ , Is there any way to use integral for Evaluation ?

Answer 1

It's telescopic. By Sophie Germain's identity $$n^4+4 = (n^2+2)^2-(2n)^2 = (n^2-2n+2)(n^2+2n+2),$$ so $$ \frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2} = \frac{1}{f(n)}-\frac{1}{f(n+2)} = \frac{4n}{n^4+4} $$ and $$ \sum_{n\geq 1}\frac{n}{n^4+4}=\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{f(n)}-\frac{1}{f(n+2)}\right) = \frac{1}{4}\left(\frac{1}{f(1)}+\frac{1}{f(2)}\right)=\frac{3}{8}. $$

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Through the inverse Laplace transform:

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{n}{n^4+4}&=&\lim_{\varepsilon\to 0^+}\frac{1}{2}\sum_{n\geq 1}\int_{0}^{+\infty}\sin(s)\sinh(s)e^{-(n+\varepsilon)s}\,ds\\&=&\frac{1}{4}\lim_{\varepsilon\to 0^+}\int_{0}^{+\infty}(1+e^{-s})\sin(s)e^{-\varepsilon s}\,ds\\&=&\frac{1}{4}\left(1+\frac{1}{2}\right)\end{eqnarray*}$$ we reach the same conclusion.