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?
A power series problem, find ROC
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real-analysis
power-series
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0$|\sin x|\leq|x|$ is a good bound when $x$ is small, but for large $x$ one should use $|\sin x|\leq1$. – 2015-02-01
2 Answers
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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.