Evaluate $\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$

Question

Evaluate $$\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$$

Does anyone have any smart ideas how to evaluate such a sum? I know one solution with complex numbers and complex analysis but I'm looking for some more smart or sophisticated methods.

Answer 1

I would not say that it is elegant, but:

The form $n^4+1$ in the denominator suggests that one should be able to get this series by expanding a combination of a hyperbolic and trigonometric function in a Fourier series.

Indeed, after some trial and error, the following function seems to work:

$$ \begin{gathered} \left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)-\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)\right)\cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right) \\ + \left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)+\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)\right)\sin \left(\frac{x}{\sqrt{2}}\right) \sinh \left(\frac{x}{\sqrt{2}}\right) \end{gathered} $$

It is even, and its cosine coefficients are $$ \frac{\sqrt{2}\bigl(\cos(\sqrt{2}\pi)-\cosh(\sqrt{2}\pi)\bigr)(-1)^{n+1} n^2}{\pi(1+n^4)},\quad n\geq 1. $$ (The zero:th coefficient is also zero). Evaluating at $x=0$ (the series converges pointwise there) gives $$ \sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n^2}{1+n^4}= \frac{\pi\left(\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)-\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)\right)}{\sqrt{2}\bigl(\cosh(\sqrt{2}\pi)-\cos(\sqrt{2}\pi)\bigr)}\approx 0.336. $$

Answer 2

Partial elements methods should works but are boring.

Let's try smthg else : having $n^4$ and $n$ of the same parity

$$ (-1)^{n^4}=(-1)^n $$ so $$ u_n= \int_0^1 n^2(-t)^{n^4} $$

So you can begin to have fun with it (I'm searching as I'm writting )