Value of $\sum_{k=2}^\infty \frac{1}{k^4+1}$

Question

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? $$\sum_{k=2}^\infty \frac{1}{k^4+1}$$

Answer 1

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 !

Answer 2

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 $$ \sum_{k\geq 2}\frac{1}{k^4+1}=-1+\frac{\pi}{2\sqrt{2}}\cdot\frac{\sinh(\pi\sqrt{2})+\sin(\pi\sqrt{2})}{\cosh(\pi\sqrt{2})-\cos(\pi\sqrt{2})}.\tag{2} $$