What is the closed form for $S=\sum_{n=1}^{\infty} \frac{\sin ({n})}{n!}$?
How do we find the following sum (closed form)?
$$S=\sum_{n=1}^{\infty} \frac{\sin ({n})}{n!}$$
Use the identity $\cos(n)+i\sin(n)=(\cos(1)+i\sin(1))^n$: $$\begin{aligned} \sum_{n=1}^\infty\frac{\sin(n)}{n!}&=\Im\sum_{n=1}^\infty\frac{\cos(n)+i\sin(n)}{n!}\\ &=\Im\sum_{n=1}^\infty\frac{(\cos(1)+i\sin(1))^n}{n!}\\ &=\Im e^{\cos(1)+i\sin(1)}\\ &=e^{\cos(1)}\sin(\sin(1)). \end{aligned}$$
$$\sum_{n=1}^\infty \frac{\sin(n)}{n!} = \Im\sum\frac{e^{in}}{n!}=\Im e^{e^i}=\Im e^{\cos(1)}e^{i\sin(1)}=e^{\cos(1)}\sin(\sin(1)).$$