Find the radius of convergence of summesion (n=0 to ∞)(an*x^n), where an=[sin(n!)/n!] and ao=0?

Question

 found a question like this and there are four options

1. R>=1
2. R>=2*pi
3. R<=4*pi
4. R<=pi, where R denotes radius of convergence. I've tried hard but cant get it how to show in equality. Also there is question in my mind that among four options if the series is not converge at the end point can we discard that option. I'm not sure abut this since we know if anywhere "<" holds we can say "<=" holds, but the converse is'nt not true.

Please help to find the solution. Please answer it in the answer section rather comment. Thanks in advance.

I repost this because dont know why my previous post is locked.

Answer 1

The Cauchy-Hadamard formula gives $$0 \leq R^{-1} = \limsup_{n \to \infty} \sqrt[n]{\left| \frac{\sin n!}{n!} \right|} \leq \limsup_{n \to \infty} \sqrt[n]{ \frac{1}{n!} } = 0. $$ Hence the radius of convergence is infinite.