Radius of convergence of power series $\sum_{n=1}^{\infty}{\frac{\sin n!}{n!}} {x^n}$

Question

The power series $\displaystyle\sum_{n=1}^{\infty}{\frac{\sin n!}{n!}} {x^n}$ has radius of convergence $R$, then

1. $R\geq1$

2. $R\geq e$

3. $R\geq2e$

4. All are correct

I wanted to know how will get the answer in the form of $e$.

Answer 1

Expanded comment.

You can prove your series converges for any $x$, by writing the rest:

$$\left|\sum_{n=m+1}^{\infty} \frac{\sin n!}{n!}x^n\right| \leq \sum_{n=m+1}^{\infty} \left|\frac{\sin n!}{n!}x^n\right| \leq \sum_{n=m+1}^{\infty} \left|\frac{x^n}{n!}\right|$$

And the last is the rest of the series $\sum_{n=0}^{\infty} \frac{|x|^n}{n!}$, which has infinite radius of convergence, as it's the series of $\exp |x|$, whose radius you should know (otherwise, prove it by the ratio test). So the rest converges to $0$ as $m\rightarrow\infty$. Hence Your series converges for any $x$, and it has also infinite radius of convergence.