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I know that the value of this limit $$\sum_{k=2}^\infty \frac{1}{k^2+1}$$ is $\frac{\pi\coth\pi−1}{2}$

Is there a similar value of this limit? $$\sum_{k=2}^\infty \frac{1}{k^4+1}$$

Henrik supports the community
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Mister Set
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    Do you have any reason to expect the result to be nice? It clearly converges and can be approximated well enough by adding the first few terms together, and can be made more accurate to any degree of desired accuracy by adding more terms. – JMoravitz Jun 18 '18 at 19:48
  • I have edited the question so it should be clear why I believe that there is nice limit – Mister Set Jun 18 '18 at 19:53

2 Answers2

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The Poisson summation formula used in the answer to your previous question applies in this case, too.
We have $$ \sum_{k\in\mathbb{Z}}\frac{1}{k^4+1} = \frac{\pi}{\sqrt{2}}\cdot\frac{\sinh(\pi\sqrt{2})+\sin(\pi\sqrt{2})}{\cosh(\pi\sqrt{2})-\cos(\pi\sqrt{2})}\tag{1}$$ hence $$ \sum_{k\geq 2}\frac{1}{k^4+1}=-1+\frac{\pi}{2\sqrt{2}}\cdot\frac{\sinh(\pi\sqrt{2})+\sin(\pi\sqrt{2})}{\cosh(\pi\sqrt{2})-\cos(\pi\sqrt{2})}.\tag{2} $$

Jack D'Aurizio
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Yes you could compare to a integral like since $f$ decrease:

Define $$ f\triangleq \dfrac{1}{1+x^4} $$

Then evaluate $$ \int \dfrac{1}{1+x^4}dx$$

Decomposition gives:

$$\dfrac{1}{1+x^4}=-\frac{1}{\sqrt{8}}[\dfrac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}+\dfrac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}] $$

Then use canonic form of denominator then you see the light !

Pagode
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