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Find the radius of convergence of the power series $\sum_{n=0}^{\infty} a_n x^n$ where $a_n={{\sin (n!)}\over {n!}}.$

Now using the ratio test $$R=\lim_{n\rightarrow \infty}\left|{{a_n}\over {a_{n+1}}}\right|\\=\lim_{n\rightarrow \infty}\left|n\cdot{\sin(n!)\over \sin((n+1)!)}\right|$$ Now, $n\rightarrow \infty$ but the limit of ${\sin(n!)\over \sin((n+1)!)}$ is not known to me. If in degrees, I could say it converges to $1$ as $360$ devides every integer of form $n!,\forall n\ge 360.$ and the $R$ would be $\infty.$ But as the questions says, I have to find this in radians. The options are $1.R\ge 1, 2.R\ge 2\pi, R\le 4\pi , R\ge \pi.$ Please help. thanks.

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If you use the root test; since $|\sin(n!)| < 1$ and $n!^{1/n} \to n/e$, $|{{\sin (n!)}\over {n!}}|^{1/n} <\frac{1}{n/e} \to 0 $ so the series converges everywhere.

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    $\begingroup$ to me, it is not obvious that $n!^{1/n} \to n/e $? $\endgroup$ – ILoveMath Nov 25 '16 at 6:55
  • $\begingroup$ Is this based on Stirling's approximation? $\endgroup$ – Henricus V. Nov 25 '16 at 7:33
  • $\begingroup$ Ridiculous down-vote. +1 Indeed this uses Stirling's approximation. $\endgroup$ – mathcounterexamples.net Nov 25 '16 at 7:58
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Hint: Compare it to Taylor's series of $e^x$ function , and that series converges everywhere.

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