Prove: $\sum_{n=0}^\infty \frac{a_n}{n!}x^n$ converges for every $x$ if $\sum_{n=0}^\infty a_nx^n$ has radius of convergence $R>0$

Question

$\sum_{n=0}^\infty a_nx^n$ has radius of convergence $R>0$

Prove: $\sum_{n=0}^\infty \frac{a_n}{n!}x^n$ converges for every $x$.

Also, I don't understand why $R < 1$ implies $$\lim_{n \to \infty} a_n \neq 0$$

If the limit exists the solution would be easy but I can't be sure it exist.

Answer 1

Hint: $\sum |a_n| (\frac R 2)^{n} <\infty$. This implies that $|a_n| (\frac R 2)^{n} $ is bounded. Suppose $|a_n| (\frac R 2)^{n} \leq C$. Then $\sum \frac {a_nx^{n}} {n!}$ is dominated by $C \sum \frac {(2|x|/R)^{n}} {n!}$.