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The task is to determine if the series converges absolutely, conditionally or doesn't converge at all.

$$\sum_{k=1}^\infty \frac{\sin(k)}{k!}$$

I have tried solving it with D'Alembert test and comparison test method. No luck. We haven't covered integration of

$$\int_1^\infty \frac{\sin(k)}{k}$$

I am stuck. Please give me a hint how to solve it.

Thank you for your attention!

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    $\begingroup$ $|\sin k/k!|\le1/k!$. $\endgroup$ – David Mitra Feb 23 '14 at 0:34
  • $\begingroup$ but 1/k! doesn't converge? $\endgroup$ – Anton Popov Feb 23 '14 at 0:35
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    $\begingroup$ exp is very useful in this context. $\endgroup$ – Max Feb 23 '14 at 0:36
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    $\begingroup$ $\sum 1/k!$ indeed converges, as the Ratio test will show. Or, compare with $\sum 1/2^k$. $\endgroup$ – David Mitra Feb 23 '14 at 0:36
  • $\begingroup$ sorry, I found my crucial mistake. I've tested 1/k! with D'Alembert and turned the fraction upside down. Thank you! $\endgroup$ – Anton Popov Feb 23 '14 at 0:38
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First we look at the absolutevalue of the series. You know that the $\sum_{n=1}^{ \infty} \dfrac {1}{n^2}$ converges, so to prove convergence (using the comparison test) it is enough to prove that for large enough $n$, $n(n-1)(n-2)>n^2$

$n(n^2-3n+2)>n^2$

$n^2-4n+2>0$ This is clearly true for large enough $n$.

If the absolute value of the series converges, then the original one does also.

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Hint: Think about $e^1 {}{}{}{}{}{}{}{}{}{}{}{}{}$

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  • $\begingroup$ indeed. I was going in circles with a silly initial mistake. $\endgroup$ – Anton Popov Feb 23 '14 at 0:42
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Taking into account MPW's answer, you can easily establish that $$\sum_{k=1}^\infty \frac{\sin(k)}{k!}=\frac{1}{2} i \left(e^{e^{-i}}-e^{e^i}\right)=\sin (\sin (1)) e^{\cos (1)}$$

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  • $\begingroup$ @nbubis. I appreciate your opinion ! Thanks and cheers. $\endgroup$ – Claude Leibovici Feb 23 '14 at 6:48
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Note: It is a year too late to answer, but I can't help it.

Since $|\sin k| < 1$, we have $$\sum \frac {|\sin k|} {k!} < \sum \frac {1} {k!} = e - 1;$$ the series converges absolutely.

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More than a convergence test: we can get an exteem of its value: $$ -e+1=-\sum_{k\ge1}\frac{1}{k!}\le\sum_{k\ge1}\frac{\sin k}{k!} \le\sum_{k\ge1}\frac{1}{k!}=e-1\;\;. $$

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