# Find the sum of the series $\sum_{n=1}^{\infty}\frac{\cos nx}{n!}$, $\sum_{n=1}^{\infty}\frac{\sin nx}{n!}$

Find the sums of the series: $$\sum_{n=1}^{\infty}\frac{\cos nx}{n!},\ \ \ \sum_{n=1}^{\infty}\frac{\sin nx}{n!}$$

I did something like this: \begin{aligned} &\sum_{n=1}^{\infty}\frac{\cos nx}{n!}+i\sum_{n=1}^{\infty}\frac{\sin nx}{n!}=\sum_{n=1}^\infty\frac{(\cos x+i\sin x)^n}{n!}=e^{\cos x +i\sin x}-1=\\ &=e^{\cos x}\cdot e^{i\sin x}-1=e^{\cos x}(\cos(\sin x)+i\sin(\sin x))-1\Rightarrow\\ &\Rightarrow \sum_{n=1}^{\infty}\frac{\cos nx}{n!}=e^{\cos x}\cos(\sin x)-1,\ \ \sum_{n=1}^{\infty}\frac{\sin nx}{n!}=e^{\cos x}\sin(\sin x) \end{aligned} However, the answer section says that $$\sum_{n=1}^{\infty}\frac{\cos nx}{n!}=e^{\cos x}\cos(\sin x)$$, and I have no idea where the number $$1$$ got from there.

• If summation starts at $n=1$ then your result is correct (as you can see by setting $x=0$). Commented Jan 30, 2020 at 8:43

Just check the two answers when $$x=0$$. You will see that the given answer is wrong. Your answer is the right one.