calculate $\sum_{k=0}^\infty \frac{1}{k!}e^{ikx}$

Question

I am new to sequences and series and I was wondering how to approach such a problem. I know that $|\frac{1}{k!}e^{ikx}| = \frac{1}{k!}$ which gets bounded by $\frac{1}{2^k}$ when $k$ reaches $4$ so $\sum_{k=0}^\infty |\frac{1}{k!}e^{ikx}|$ converges and so does the sum without absolute value. I am not able to solve this sum, I tried to find some tricks but couldn't find any. How do I calculate this sum and is there a general approach or basic things I can try to calculate other sums in the future?

Answer 1

It is an exponential series, whose variable is itself an exponential, such that $$ \sum_{k=0}^\infty\frac{(e^{ix})^k}{k!} = e^{e^{ix}} = e^{\cos x + i\sin x} = e^{\cos x}(\cos(\sin x) + i\sin(\sin x)) $$