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From this answer, I was very interested to learn the following.

Let $x_n$ be the $n$-th positive root of the equation $\csc x=x$. Then $$\sum_{n\ge1}\frac{1}{x_n^2}=1,$$ and, setting $s(k)=\sum_{n\ge1}x_n^{-k}$, we have $$\sum_{k\ge1}s(2k)x^{2k}=\frac{x}{2}\cdot\frac{1+x\cot x}{\csc x-x}.\tag 1$$

This result was very surprising and I've never seen anything like it before. The proof was discussed rather briefly in the comments and apparently it can be shown via a contour integral, but I've never done anything like that before so I have no idea how. Could I have some help proving $(1)$? Thanks! :)

clathratus
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1 Answers1

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The $x_n$ are the zero of $z \sin z-1$ (all simple). The sequence is almost periodic (because the large zeros of $\sin z - 1/z$ are close to $\pi n$). So $$f(z)=\frac{(z\sin z -1)'}{z\sin z -1}-\sum_n (\frac1{z-x_n}+\frac{1}{x_n})$$ converges, it is entire and $f'$ is bounded and vanishes at $\infty$ so $f'=0$ and $f$ is constant, since $f$ is odd then $f=0$.

For $|z|<|x_1|$ the smallest zero $$g(z)=\frac{(z\sin z -1)'}{z\sin z -1}=\sum_n (\frac1{z-x_n}+\frac{1}{x_n})$$ is analytic and $$g^{(2k)}(0)=0, \qquad g^{(2k+1)}(0)=- (2k+1)!\sum_{n} \frac1{x_n^{2k+2}}$$ whence $$g(z)=-\sum_{k=0}^\infty z^{2k+1}\sum_{n} \frac1{x_n^{2k+2}}$$

reuns
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