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Is $$\sum_{n=1}^{\infty}\frac{1}{n\sin n}$$ convergent or not?

WuKong
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    By Hurwitz theorem for rational approximation of irrational numbers, $|\pi - n/m| < \frac{1}{\sqrt{5}m^2}$ for infinitely many $n,m$. This implies $\frac{1}{|\sin n|} > Cn$ for infinitely many $n$ and, thus $\frac{1}{n \sin n}$ does not converge to $0$ so the series diverges. – RRL Apr 12 '19 at 04:36
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    @YuiTo Cheng: This is a great problem but not the best posed question for the site with no context/effort. – RRL Apr 12 '19 at 04:55
  • Can somebody give an elementary proof for $n\sin n \nrightarrow \infty$ as $n \to \infty$? – WuKong Apr 12 '19 at 04:57
  • @RRL what's $C_n$? Could you elaborate your proof,please? – WuKong Apr 12 '19 at 05:02
  • @mengdie1928: Ok - below - I don't know of anything more elementary. – RRL Apr 12 '19 at 05:08
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    This type of series question -- with other powers $n^a (\sin n)^b$ as well -- is always about to what extent $\sin n$ is repeatedly close to $0$ for all $n \in \mathbb{N}$. There are series of this type where the convergence is unknown. Read about the Flint Hills problem or look on this site. Not enough is known about the irrationality measure of $\pi$. – RRL Apr 12 '19 at 05:13

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We have $|\sin n| = |\sin(m\pi -n)| < |m\pi - n| < \frac{1}{\sqrt{5}m} =\frac{1}{\sqrt{5}n} \frac{n}{m} $ for infinitely many $n,m$, where the inequality follows from the Hurwitz rational approximation $|m\pi - n| < \frac{1}{\sqrt{5}m}$.

Again from the approximation $\frac{n}{m} < \pi + \frac{1}{\sqrt{5}m^2 } < \pi + 1$.

Thus, $|\sin n| < \frac{\pi +1}{\sqrt{5}}\frac{1}{n}$ and $(n|\sin n|)^{-1}> \frac{\sqrt{5}}{\pi +1}$for infinitely many $n$. Hence, there is a subsequence of $\{(n\sin n)^{-1}\}$ that does not converge to $0$ and the series diverges.

RRL
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