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I don't know if my reasoning is right on this exercise: If the power series $\sum a_n z^n$ has radius of convergence $R$, which is the radius of convergence of the series $\sum a_n^2 z^n$ and $\sum a_n z^{2n}$?

For the first one I am thinking: $\displaystyle \lim_{n \to \infty } \left|\dfrac{a^2_{n+1}}{a^2_n}\right|=\lim_{n \to \infty } \left|\dfrac{a_{n+1}}{a_n}\right|^2$= $R^2$.

And for the second, I believe it must have the same radius $R$, because it has the same coefficients, but we are only taking the pair powers, is this right?

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  • $\begingroup$ You need a tool that does not assume every $a_n$ is nonzero. Which ones do you know? $\endgroup$ – Did May 30 '13 at 15:53
  • $\begingroup$ @Did Isn't it trivial to assume that all $a_n$ are nonzero? If some $a_n$ is zero, it does not contribute to the sum, so you may concatenate the sequence. $\endgroup$ – alex.jordan May 30 '13 at 15:57
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    $\begingroup$ The second series replaces $z$ by $z^2$. Formerly you required $|z|<R$. Now you require $|z^2|<R$, equivalent to $|z|<\sqrt{R}$. $\endgroup$ – alex.jordan May 30 '13 at 15:59
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    $\begingroup$ If $R$ is a radius of convergence of $\sum a_n z^n$ does not mean that $\lim\limits_{n\to\infty}\left|\frac{a_n}{a_{n+1}} \right|$ exists at all. $\endgroup$ – M. Strochyk May 30 '13 at 16:03
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    $\begingroup$ You could use $\limsup$ of the $n$-th root, since that gives an if and only if criterion, though the answer of Did is I think more concrete. $\endgroup$ – André Nicolas May 30 '13 at 16:31
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Hint: The radius of convergence $R$ of the series $\sum\limits_na_nz^n$ is characterized by the fact that $|a_n|r^n\to0$ when $r\lt R$ and $|a_n|r^n\not\to0$ when $r\gt R$ (note how one carefully avoids to specify what happens when $r=R$, with reason since several different behaviours are possible).

Thus, $|a_n^2|r^n=(|a_n|(\sqrt{r})^n)^2\to0$ if $\sqrt{r}\lt R$ and $|a_n^2|r^n\not\to0$ if $\sqrt{r}\gt R$. These conditions can be rewritten as $r\lt R^2$ and $r\gt R^2$, respectively, hence the radius of convergence of $\sum\limits_na_n^2z^n$ is $R^2$.

Surely you can proceed likewise for $\sum\limits_na_nz^{2n}$.

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