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The question is closely linked with Convergence of the sequence $\frac{1}{n\sin(n)}$. I am afraid if I am not interpreting it wrong (read it for the first time when I was asked to do the exercise that I mentioned above). Also the comment box of the above mentioned question gave a helpful info. It was mentioned that if we refer to the sequence $\frac{1}{n^u\sin^vn}$, then it converges to $0$ as soon as $1+\frac{u}{v}>7.6063$. The exercise that I mentioned above does not hold it fine. So if it doesn't hold then the sequence will not get close to $0$. Sine the initial criteria for the convergence of a series doesn't hold, therefore the series will diverge as well. To write its theoretical proof, the same situation as in above link mentioned will happen. To which non-zero irrational number the sequence in my question will converge(if it do)? Correct if I misinterpreted.

Aastha Choudhary
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    The result says that if $1+\frac{u}{v}>7.6063$, then the sequence converges. But it doesn't claim that this is an if and only if, so trying to argue the opposite statement (if the inequality fails, the sequence diverges) is not valid. Indeed we should expect convergence whenever $1+\frac{u}{v}>2$. So the series in question should converge, but I doubt anyone knows how to formally prove it. – Wojowu Jul 09 '22 at 15:20
  • Thanks for clarifying @Wojowu. – Aastha Choudhary Jul 09 '22 at 16:08

1 Answers1

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It is not known whether this series converges, nor even whether $$ \lim_{n\rightarrow\infty} \frac1{n^4 \sin(n) } = 0 $$ Corollary 3 of Alekseyev (cited in the answer you link) implies that $\lim_{n\rightarrow\infty} \frac1{n^4 \sin(n) } = 0$ would imply $\mu(\pi) \le 5$, where $\mu(\cdot)$ is the irrationality measure. But currently, all we know is that $$ 2\le \mu(\pi)\le 7.1032... $$ If $\mu(\pi) < 4$, then $\sum\limits_{n=0}^\infty \frac1{n^4 \sin(n)}$ converges absolutely, whereas if $\mu(\pi) > 5$ then the sum diverges because the summands do not go to $0$. If $\mu(\pi)\in[4,5)$, then I suppose the series will converge but not absolutely, as it is basically like an alternating series.

Dark Malthorp
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    Just for completeness let me say that it is generally expected that $\mu(\pi)=2$, so the series should indeed converge absolutely. – Wojowu Jul 09 '22 at 16:10