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I want to calculate the sum of $$\sum_{n\geq 0}\frac{\sin n}{n!}.$$

I think I am supposed to use the Taylor polynomial of $\ e^x$ but I don't know how to solve it.

Thanks for your help.

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  • $\begingroup$ @graydad I see it's been edited. $\endgroup$ – Akiva Weinberger Jan 23 '15 at 0:05
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Since $\sin x =\Im(e^{ix})$, we have: $$ \sum_{n\geq 0}\frac{\sin n}{n!} = \Im\left(e^{e^i}\right) = e^{\cos 1}\sin(\sin 1).$$

[$\Im(z)$ (also written "$\operatorname{Im}(z)$") means the imaginary part of $z$.]

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    $\begingroup$ Perhaps it's better to write \text{Im} , as that notation for "Imaginary" sucks...and many just don't understand it. $\endgroup$ – Timbuc Jan 22 '15 at 23:45
  • $\begingroup$ @JackD'Aurizio thanks for your answer. Could you show me the second equality using baby steps please? $\endgroup$ – user1118686 Jan 22 '15 at 23:55
  • $\begingroup$ @user1118686 You need to know this formula first: $e^{ix}=\cos(x)+i\sin(x)$. Do you? $\endgroup$ – Akiva Weinberger Jan 22 '15 at 23:55
  • $\begingroup$ @user1118686: we have that $\Im$ is a linear operator and for any $z$, $$\sum_{n\geq 0}\frac{z^n}{n!}=e^{z},$$ so we plug in $z=e^{i}$ and take the imaginary part. $\endgroup$ – Jack D'Aurizio Jan 22 '15 at 23:57
  • $\begingroup$ $\displaystyle\sum_{n\geq0}\frac{\sin n}{n!}=\sum_{n\geq0}\frac{\Im(e^{ix})}{n!}= \Im\sum_{n\geq0}\frac{e^{ix}}{n!}=\Im\left(e^{e^i}\right)=e^{\cos 1}\sin(\sin 1)$ $\endgroup$ – Akiva Weinberger Jan 22 '15 at 23:59

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