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Well, I am always wrong in calculation of Radius of Convergence $R$ of a Power series. Would anyone help me to find this one? $\sum\limits_{n=0}^{\infty}a_nx^n$ where $a_0=0$ and $a_n= \sin(n!)/n!$, I guess $R\ge 1$, is it?

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Do you know a good bound for $|\sin n!|$? – Gerry Myerson Jun 2 '12 at 12:34
the fastest way to reason here is to compare this with the power series of the exponential function (which is $\infty$) – user20266 Jun 2 '12 at 12:34
gerry,I know $|sinx|\le |x|$, Thomas I did not get your point. – Taxi Driver Jun 2 '12 at 12:36
@Mex There is a much better (constant) bound. – Davide Giraudo Jun 2 '12 at 12:47
so here It will be $n!$ – Taxi Driver Jun 2 '12 at 12:50

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Wouldn't $\,|\sin n!|\leq 1\,$ be a sharper inequality? With it you can even try absolute convergence for any given $\,x\in\mathbb{R}\,$ and get a pretty big ROC...

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The formula for the radius of convergence of a power series is $$ \frac1R=\limsup_{n\to\infty}a_n^{1/n}\tag{1} $$ That and Stirling's Approximation should answer your question.

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