There are given two trigonometric series:
$$\sum_{n=0}^{\infty}\frac {cos(nx)}{n!}\, and\, \sum_{n=0}^{\infty}\frac {sin(nx)}{n!} $$
1) Show that the two series are uniform convergent on the interval R. The sum functions for the two series are defined as $f$ and $g$ respectively. And explain why $f$ and $g$ are continuous and periodic with $2\pi$.
2) Determine $f(x)$ and $g(x)$ for every $x\in \mathbb{R}$ by computing $f(x) +ig(x)$ (Hint: it can be used that for the complex exponential function $e^z$ it is true that: $$ e^z = \sum_{n=0}^{\infty}\frac {z^n}{n!}$$
1) I have found a majorant series for $f$ and $g$: $$\sum_{n=0}^{\infty}\frac {\lvert cos(nx)\rvert}{n!} \le\sum_{n=0}^{\infty}\frac {1}{n!}$$ $$\sum_{n=0}^{\infty}\frac {\lvert sin(nx)\rvert}{n!} \le\sum_{n=0}^{\infty}\frac {1}{n!}$$ and by performing the ratio text on the majorant series I know that it converges and therefore the two series are uniform convergent. Since the two series are uniform convergent on the interval R, this implies that they are also continuous. And $sin(x+2\pi)=sin(x)$ and $cos(x+2\pi)=cos(x)$ Therefore, $f$ and $g$ are periodic with $2\pi$.
2) I am not sure how to determine $f(x)$ and $g(x)$ by computing $f(x)+ig(x)$.
Hint. From $$ e^z = \sum_{n=0}^{\infty}\frac {z^n}{n!}$$ one gets $$ e^{e^{ix}} = \sum_{n=0}^{\infty}\frac {e^{inx}}{n!}=\sum_{n=0}^{\infty}\frac {\cos(nx)+i\sin(nx)}{n!}= \sum_{n=0}^{\infty}\frac {\cos(nx)}{n!}+ i\sum_{n=0}^{\infty}\frac {\sin(nx)}{n!}. $$ Then find, for $x \in \mathbb{R}$, $$ \text{Re}\left( e^{e^{ix}}\right),\quad \text{Im}\left( e^{e^{ix}}\right). $$