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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.

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    $\begingroup$ If you are looking for a real analysis solution, then why are the tags complex-analysis and complex-numbers appropriate? $\endgroup$ – Mark Viola May 14 '18 at 16:37
  • $\begingroup$ not necessarily for a real analysis! I haven’t said that. But maybe there is a nice solution in complex analysis but not that I already know $\endgroup$ – Karagum May 14 '18 at 16:38
  • $\begingroup$ Well, what one do you already know? $\endgroup$ – Mark Viola May 14 '18 at 16:39
  • $\begingroup$ Let $f(z)=\frac{z^2}{(n^4+1) \sin( \pi z)}$. We found poles, residuums of $f(z)$ and use Cauchy's residue formula $\endgroup$ – Karagum May 14 '18 at 16:42
  • $\begingroup$ Yes indeed, sorry, it should be $z^4+1$, not $n^4+1$ $\endgroup$ – Karagum May 14 '18 at 16:51
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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 $$ \sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n^2}{1+n^4}= \frac{\pi\left(\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)-\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)\right)}{\sqrt{2}\bigl(\cosh(\sqrt{2}\pi)-\cos(\sqrt{2}\pi)\bigr)}\approx 0.336. $$

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  • $\begingroup$ That seems like quite a bit of trial and error and a bit contrived in a way. But if does give the correct result so (+1) $\endgroup$ – Mark Viola May 14 '18 at 18:08
  • $\begingroup$ @MarkViola Well, it was not so difficult, just make the correct ansatz and find the appropriate constants. But I agree it is not really elegant, smart or sophisticated. Thanks for the upvote. $\endgroup$ – mickep May 14 '18 at 18:19
  • $\begingroup$ @mickep Could you please give more explanations and steps of this? It seems interesting but I cannot undestand any single step etc $\endgroup$ – Karagum May 14 '18 at 19:57
  • $\begingroup$ Honestly, when you say that you don't understand a single step I have no clue where to start. $\endgroup$ – mickep May 14 '18 at 20:09
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    $\begingroup$ If you just want formulas you could for example look in Table of integrals, series, and products by Gradshteyn and Ryzhik. If you want to understand the theory of Fourier series, then I recommend the book Fourier analysis and its applications by Folland. $\endgroup$ – mickep May 15 '18 at 17:27
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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 )

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