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I know there is a closed form for $$\sum_{n=1}^{\infty}\frac{\sin n}{n!},$$ and I am wondering if there is also a closed form for $$\sum_{n=1}^{\infty}\frac{\csc n}{n!}.$$ My Attempt: $$\sum_{n=1}^{\infty}\frac{\csc n}{n!}$$ $$= \sum_{n=1}^{\infty}{\frac{2i}{(e^{in}-e^{-in})n!}}$$ $$ = 2i\sum_{n=1}^{\infty}{\frac{1}{\left(\frac{e^{2in}}{e^{in}}-\frac{1}{e^{in}}\right)n!}}$$ $$= 2i\sum_{n=1}^{\infty}{\frac{e^{in}}{(e^{2in}-1)n!}}$$ I know the taylor series $$ e^x = \sum_{n=1}^{\infty}{\frac{x^{n}}{n!}},$$ but I don't know how to continue the problem anymore.

To see whether the series converges, please visit Convergency of $\sum_{n=1}^{\infty}\frac{\csc(n)}{n!}$

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Larry
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    I think a closed form is unlikely. Even to prove the sum converges is not obvious (relying on an estimate of how well $\pi$ can be approximated by rationals). – Robert Israel Sep 27 '18 at 22:52
  • I think so too, but maybe there is one. I also added a link that is about the convergency of the series. – Larry Sep 27 '18 at 23:26

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