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How do we find the following sum (closed form)?

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

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  • $\begingroup$ Use (one of) Euler's formulae. $\endgroup$ – Daniel Fischer Jul 17 '14 at 16:09
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    $\begingroup$ $\Im e^{e^i} = e^{\cos 1}\sin(\sin 1)$ $\endgroup$ – achille hui Jul 17 '14 at 16:11
  • $\begingroup$ If you have a new question, you should click on "ask question" and ask a new one; not edit the previous one. $\endgroup$ – Najib Idrissi Jul 18 '14 at 8:26
  • $\begingroup$ @NajibIdrissi This isn't a new question..i know how this is done..i shared it because the answer is so elegant.. $\endgroup$ – pkwssis Jul 18 '14 at 8:29
  • $\begingroup$ Well then, if this is an answer, you should post it as an answer to your question, and probably add more details... $\endgroup$ – Najib Idrissi Jul 18 '14 at 8:31
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Use the identity $\cos(n)+i\sin(n)=(\cos(1)+i\sin(1))^n$: $$\begin{aligned} \sum_{n=1}^\infty\frac{\sin(n)}{n!}&=\Im\sum_{n=1}^\infty\frac{\cos(n)+i\sin(n)}{n!}\\ &=\Im\sum_{n=1}^\infty\frac{(\cos(1)+i\sin(1))^n}{n!}\\ &=\Im e^{\cos(1)+i\sin(1)}\\ &=e^{\cos(1)}\sin(\sin(1)). \end{aligned}$$

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$$\sum_{n=1}^\infty \frac{\sin(n)}{n!} = \Im\sum\frac{e^{in}}{n!}=\Im e^{e^i}=\Im e^{\cos(1)}e^{i\sin(1)}=e^{\cos(1)}\sin(\sin(1)).$$

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    $\begingroup$ How is this different from the other answer? $\endgroup$ – Artem Jul 21 '16 at 2:18
  • $\begingroup$ The symbols are different. Perhaps slightly more succinct. I prefer these steps myself. $\endgroup$ – Pixel Jul 21 '16 at 8:23

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