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I recently came across the following summation: $$\sum_{n=0}^\infty\zeta(2+2n)x^{2n}=\frac{1-\pi x\cot(\pi x)}{2x^2},$$ where $\zeta$ is the Riemann zeta function. Wolfram MathWorld cites Bailey et. al. 2006, but I can't seem to find that source in the references. Does anybody know where this reference is located (or any other place to find the evaluation of this sum)?

csch2
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1 Answers1

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In this MSE question the OP derives to a very similar expression $$x\cot x = 1-2\sum_{n=1}^{\infty}\zeta(2n)\frac{x^{2n}}{\pi^{2n}}.$$ To get to this result he starts with the Weiertrass product of the sine. So, how can we get from this expression to yours? First of all we can bring the $1$ as well as the $-2$ to the left hand side. $$\frac{1-x\cot x}{2} = \sum_{n=1}^{\infty}\zeta(2n)\frac{x^{2n}}{\pi^{2n}}.$$ Now substitute $x\to\pi x$. This yields $$\frac{1-\pi x\cot \pi x}{2} =\sum_{n=1}^{\infty}\zeta(2n)\frac{(\pi x)^{2n}}{\pi^{2n}}=\sum_{n=1}^{\infty}\zeta(2n)x^{2n}.$$

Start from $n=0$ instead from $n=1$, by writing every $n$ as $n+1$. $$\sum_{n=1}^{\infty}\zeta(2n)x^{2n}=\sum_{n=0}^{\infty}\zeta(2(n+1))x^{2(n+1)} =x^2\sum_{n=0}^{\infty}\zeta(2n+2)x^{2n}.$$ Remember what the sum was equal to and divide both sides by $x^2$ $${\frac{1-\pi x\cot \pi x}{2x^2} = \sum_{n=0}^{\infty}\zeta(2n+2)x^{2n}}.$$

vitamin d
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  • Thanks for the very nice derivation. In case you're interested you can also use this formula to get a closed form for the sum $\sum_{n=1}^\infty[n^{2k}(x-n^2)]^{-1}$ for any natural number $k$ (which is the original application I had in mind). – csch2 Feb 24 '21 at 23:33
  • Interesting. I would like to see your proof. Do you have it as a latex document? If yes, could you post the link in here? – vitamin d Feb 24 '21 at 23:37
  • Here's the link: https://www.overleaf.com/6415664718fwggfshpsdsr. I'm not an expert in complex analysis but I'm pretty sure this is correct (Mathematica confirms it, at least for given values of $k$). – csch2 Feb 25 '21 at 00:01