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I got to this series while solving an integral by contour, I checked the result of the series with Mathematica and it is $\dfrac{1}{24}$, which gives the correct result of the integral. The problem is that I even no idea how to show that the series converges to that result.

I tried to rewrite it as a well-known converging series (like a geometric series) or to use the definition (limit of partial sums), but I couldn't manage to do either.

If it can be of any use, the series can also be written as (it won't compute with Mathematica): $$\dfrac{1}{2}\sum_{k=0}^\infty \left(2k+1\right)\left(1-\tanh\left(\dfrac{\pi}{2}\left(2k+1\right)\right)\right)$$ having used the following equality $$1-\tanh\left(x\right)=\dfrac{2}{1+e^{2x}}$$

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Siphon
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  • What's the integral? – gist076923 Mar 02 '23 at 15:04
  • @RandomVariable Most definitely, thank you very much. I couldn't find that question due to how the series was written. – Siphon Mar 02 '23 at 20:54
  • @gist076923 The integral is $\int_0^\infty \dfrac{x\sin x}{\cos x\cosh x}$

    I saw it in a video, but they didn't solve it with a contour integral. The result should be $2\pi^2$ multiplied by the series in the title.

     – Siphon Mar 02 '23 at 21:00

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