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Is there a way to assess the convergence of the following series? $$\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$$ From numerical estimations it seems to be convergent but I don't know how to prove it.

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  • $\begingroup$ Are there infinitely many summands that are $\gt\frac{1}{2n}$? I don't think it converge..Does this help?math.stackexchange.com/questions/2227574/… $\endgroup$ – Tony Ma Apr 27 '18 at 12:40
  • $\begingroup$ If there is some $\epsilon>0$ such that $\sum_{n=1}^N \sin(n!)=O(N^{1-\epsilon})$, then the series can be shown to converge by partial summation. I expect that $\sin(n!)$ is positive about as often as it is negative, and that $\sum_{n=1}^N\sin(n!)$ grows slowly, but I cannot see how to prove the necessary bound. $\endgroup$ – Julian Rosen Apr 27 '18 at 14:32
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    $\begingroup$ For almost all $\theta\in [0,2\pi)$, the series $$\sum_{n=1}^{\infty} \frac{\sin(n! \theta)}n $$ converges. However, it is a completely different story if $\theta=1$ belongs to the "almost all". $\endgroup$ – Sungjin Kim Apr 28 '18 at 18:39
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    $\begingroup$ I'm thinking use multiple angle formula for the sine and maybe Dirichlet test. $\endgroup$ – Marek Kurczynski Dec 3 '18 at 8:32
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    $\begingroup$ Can someone explain why we can't show that $\left | \frac{\sin(n!)}{n} \right | \leq \left | \frac{(-1)^n}{n} \right|$ for all $n \in \mathbb{N}$, implying the series is "bounded" by the alternating harmonic series? $\endgroup$ – Spencer Kraisler Jun 3 at 19:37

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