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I am stuck on how to prove the convergency of the series $$\sum_{n=1}^{\infty}\frac{\csc(n)}{n!}.$$ It seems like that the series converges to approximately $2.85$, but I have no idea how to show whether the series converges or not. I know that there are techniques such as the ratio test, the integral test, the $n^{th}$ term test, but I still do not know how to approach the problem.

Could you give me a hint?

Larry
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  • I don't think this converges ... $\csc(n)$ can be arbitrarily large *if you choose the integer $n$ carefully to be very close to a multiple of $\pi$. If you look at the graph of $\csc(n)$ over reals, it has asymptotes at every integer multiple of $\pi$ edit:clarified, but turns out this guess was wrong anyway! – Zubin Mukerjee Sep 25 '18 at 21:33
  • That doesn't mean it doesn't converge. Search the flint hill series for an example of an open problem. – QC_QAOA Sep 25 '18 at 21:34
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    $n!$ grows faster than any polynomial, so the given series is clearly convergent, since the irrationality measure of $\pi$ is finite. – Jack D'Aurizio Sep 25 '18 at 21:36
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    @JackD'Aurizio very interesting, I had never heard of irrationality measure. It's also interesting that the bound is still being improved even in the last year! – Zubin Mukerjee Sep 25 '18 at 21:37

1 Answers1

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$\pi\not\in\mathbb{Q}$ has a finite irrationality measure, in particular there are a finite number of $\frac{p}{q}\in\mathbb{Q}$ such that $$ \left|\pi - \frac{p}{q}\right|\leq \frac{1}{q^{10}}\quad\Longleftrightarrow\quad d(p,\pi\mathbb{Z})\leq\frac{1}{q^9} $$ and for any sufficiently large $n\in\mathbb{N}^+$ we have $$ \left|\sin(n)\right|\geq \tfrac{2}{\pi}d(n,\pi\mathbb{Z})\geq 2\pi^8\cdot\frac{1}{n^9}. $$ Since the series $\sum_{n\geq 1}\frac{n^9}{n!}$ is convergent, the series $\sum_{n\geq 1}\frac{1}{n!\sin(n)}$ is absolutely convergent.

Jack D'Aurizio
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