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Give that the radius of convergence of $ \sum\limits_{n=1}^\infty a_nz^{n}$ is $R$, find the radius of convergence $R_1$ and $R_2$ of the following series: $$\sum\limits_{n=1}^\infty \frac{a_n}{n!}z^{n}$$ $$\sum\limits_{n=1}^\infty n^ka_nz^{n}$$

My attempts in each case are to attempt to compute the following $$ R_1=[\limsup\limits_{n\rightarrow\infty}|\frac{a_n}{n!}|^{\frac{1}{n}}]^{-1}$$ $$ R_2=[\limsup\limits_{n\rightarrow\infty}|n^ka_n|^{\frac{1}{n}}]^{-1}$$ How can I proceed in taking apart the limits in order to to reach some function of $R$? I know that $R$ in this case is given by $$R=[\limsup\limits_{n\rightarrow\infty}|a_n|^{\frac{1}{n}}]^{-1}$$

Any pointers appreciated. Thankyou

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    $\begingroup$ Good exercise, in that its best solution by far, completely avoids the characterizations of the radius of convergence you recalled. One should add the condition that $R$ is not zero, then use the fact that the radius of convergence of the series $\sum\limits_na_nz^n$ being $R$ means that for every $r<R<s$, the sequence $(a_nr^n)$ goes to zero exponentially fast and the sequence $(a_ns^n)$ does not converge to zero. $\endgroup$ – Did Mar 29 '15 at 13:21
  • $\begingroup$ I understand the fact you have mentioned but I do not know how to apply it to the cases above. Why does noting that $R$ is non zero help toward finding an answer? Thanks $\endgroup$ – Trawkley Mar 29 '15 at 13:48
  • $\begingroup$ One needs to assume that $R\ne0$, otherwise the radius of convergence of the first series you ask about can be pretty much anything (note that the accepted answer seemingly falls into this trap). $\endgroup$ – Did Mar 29 '15 at 15:42
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Recalling that if $\lim_{n\rightarrow\infty}b_{n}\geq 0$ and $\limsup_{n\rightarrow\infty}a_{n}$ is finite, then $$\limsup_{n\rightarrow\infty}a_{n}b_{n}=\limsup_{n\rightarrow\infty}a_{n}\limsup_{n\rightarrow\infty}b_{n}$$ (so I'm assuming that $R\neq0$) we can note that $$\lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}=0$$ $$\lim_{n\rightarrow\infty}\sqrt[n]{n^{k}}=1$$ so $$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|\frac{a_{n}}{n!}\right|}=\limsup_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_{n}\right|}=0$$ $$\limsup_{n\rightarrow\infty}\sqrt[n]{n^{k}\left|a_{n}\right|}=\limsup_{n\rightarrow\infty}\sqrt[n]{n^{k}}\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_{n}\right|}=\frac{1}{R}.$$

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    $\begingroup$ A minor point here, but you don't actually need $a_n,b_n \ge 0$ -- the only additional hypotheses needed is $\lim_{n\to \infty} a_n \ge 0$ and $\limsup b_n$ is finite. $\endgroup$ – kobe Mar 29 '15 at 14:41
  • $\begingroup$ @kobe Fixed, thank you. $\endgroup$ – Marco Cantarini Mar 29 '15 at 14:44
  • $\begingroup$ Your post is rather ambiguous about the first series but if ever your intent with it was to show that the radius of convergence is infinite in this case, note that this is not always so (and no, the limsup you compute is not always $0$). $\endgroup$ – Did Mar 29 '15 at 15:46
  • $\begingroup$ @Did I hope it's more clear now. $\endgroup$ – Marco Cantarini Mar 29 '15 at 18:00
  • $\begingroup$ The patch is not needed for the second series. $\endgroup$ – Did Mar 29 '15 at 18:43

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