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Let power series $\sum_{n=0}^{\infty} a_nz^n$ have radius of convergence $R$. I would ask you, is it true that $\sqrt[n]{a_n} \rightarrow \frac{1}{R}$? If it is true, then power series $\sum_{n=0}^{\infty} a_n^Mz^n$ have radius of convergence $R^M$ and $\sum_{n=0}^{\infty} a_nz^{Mn}$ have radius $R^{1/M}$ ? I was using Cauchy test. Am I right?

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Consider the series $\sum a_n x^n$, where $a_n=\frac{1}{2^n}$ if $n$ is even, and $a_n=\frac{1}{3^n}$ when $n$ is odd. The radius of convergence is $2$, but the limit of $\sqrt[n]{a_n}$ does not exist.

However, if you use the Cauchy Test in its $\limsup$ form, it will get you what you want.

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  • $\begingroup$ So what can I say about radius of convergence power series $\sum_{n=0}^{\infty} a_n^Mz^n$ and $\sum_{n=0}^{\infty} a_nz^{Mn}$ if I know that $\sum_{n=0}^{\infty} a_nz^n$ have radius of convergence $R$? Of course $M \in \mathbb{N}$. $\endgroup$
    – Thomas
    Commented Mar 29, 2014 at 20:03
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    $\begingroup$ You can say what you wanted to say. My answer just pointed out that you could not use $\lim\sqrt[n]{|a_n|}=\frac{1}{R}$, since that is not true. But you can use $\limsup\sqrt[n]{|a_n|}=\frac{1}{R}$ to get $R^M$, $R^{1/M}$. $\endgroup$ Commented Mar 29, 2014 at 20:16
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If the radius of convergence of $\sum a_nz^n$ is $R$, then the radius of convergence of $\sum a_n^Mz^n$ is $R^M$.

This can be shown using root test (which is conclusive).

If $\limsup \vert a_n\rvert^{1/n}=\ell$, then clearly $\limsup \vert a_n^M\rvert^{1/n}=\ell^M$.

Note that, if $\limsup \vert a_n\rvert^{1/n}=\ell\in [0,\infty]$, then the radius of convergence $R$ of $\sum a_nz^n$ is $$ R=\left\{\begin{array}{lll} 0 & \text{if} & \ell=\infty, \\ 1/\ell & \text{if} & \ell\in(0,\infty), \\ \infty & \text{if} & \ell=0. \end{array}\right. $$

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