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if $\sum_\limits{n=1}^\infty a_n z^n$ has radius of convergence $R$, what ate the radii of convergence of $\sum_\limits{n=1}^\infty a_n z^{2n}$ and $\sum_\limits{n=1}^\infty {a_n }^2z^n$??

radius of convergence of $\sum_\limits{n=1}^\infty {a_n}^2 z^n$ is at least $R^2$. but is it exactly $R^2$?? because $lim sup(a_nb_n)\leq limsup(a_n)limsup(b_n)$ but equality may not hold.

what about other series ?

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    It stands to reason, at least, that the radius of convergence of $\sum a_nz^{2n}$ must be $\sqrt R$, since that corresponds to plugging $z^2$ into the original series. – 2017-02-08

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Note that if the radius of convergence of $\sum_{n=1}a_nz^n$ is $R$, then

$$\limsup_{n\to \infty}\sqrt[n]{a_n}=1/R$$

Therefore, we have

$$\limsup_{n\to \infty}\sqrt[n]{|a_nz^{2n}|}=|z|^2/R<1$$

whenever $|z|<\sqrt{R}$.

Furthermore, we have

$$\limsup_{n\to \infty}\sqrt[n]{|a_n^2z^{n}|}=|z|/R^2<1$$

whenever $|z|

Note that we have

$$\begin{align} \limsup_{n\to \infty}\sqrt[n]{|a_n|^2}&=\lim_{n\to \infty}\sup_{m\ge n}(|a_m|^{1/m}|a_m|^{1/m})\\\\ &=\lim_{n\to \infty}(\sup_{m\ge n}(|a_m|^{1/m})\sup_{m\ge n}(|a_m|^{1/m}))\\\\ &=\left(\limsup_{n\to \infty}\sqrt[n]{a_n}\right)^2\\\\ &=1/R^2 \end{align}$$

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    :-) Beat me to it! – 2017-02-08
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    @SimplyBeautifulArt I cannot tell you the number of times that I've been on the other side of this. – 2017-02-08