# How to compute $\sum_{n\geq 0}\frac{\sin n}{n!}$?

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.

• @graydad I see it's been edited. – Akiva Weinberger Jan 23 '15 at 0:05

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 $z$.]
• @user1118686 You need to know this formula first: $e^{ix}=\cos(x)+i\sin(x)$. Do you? – Akiva Weinberger Jan 22 '15 at 23:55
• @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. – Jack D'Aurizio Jan 22 '15 at 23:57
• $\displaystyle\sum_{n\geq0}\frac{\sin n}{n!}=\sum_{n\geq0}\frac{\Im(e^{ix})}{n!}= \Im\sum_{n\geq0}\frac{e^{ix}}{n!}=\Im\left(e^{e^i}\right)=e^{\cos 1}\sin(\sin 1)$ – Akiva Weinberger Jan 22 '15 at 23:59