What is the radius of convergence of the series $\sum_{n=0}^\infty \frac{\sin(n!)}{n!} x^n$ ?
Since in $\{a_n\}$ we have $a_n=\frac{\sin (n!)}{n!}$. So $\{a_n\}$ is a bounded sequence. Hence radius of convergence $R>1$. But how to proceed further ?
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3From the assumption $a_n$ is bounded, we can only conclude $R \geq 1$, not $R >1$. In any case try to show $\sum |a_nx^n| \leq f(|x|)$ for some function $f$ you can identify. – Brian Moehring May 25 '23 at 16:58
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6Maybe have a look into what the radius of convergence of the power series for $e^z$ is? – A rural reader May 25 '23 at 16:58
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@BrianMoehring Yes $R\geq 1$, my mistake. But how does the latter helps ? – Vishnu K S May 26 '23 at 04:28