When I was solving an exercise related to power series with complex numbers, I had the next question. I thought that was trivial, but, in fact, it isn't.
Let $\displaystyle\sum_{n=0}^{\infty} a_nz^n$ a power series (centered at 0), where for all $n\in\mathbb{N}$, $a_n\in\mathbb{C}$, such that the series converges if and only if $|z|<R$, i.e., $R$ is the radius of convergence of the series. If we take $\{a_{n_k}\}_{k\in\mathbb{N}}\subseteq\{a_n\}_{n\in\mathbb{N}}$ a subsequence of $\{a_n\}$ then, what can we say about the radius of convergence $r$ of the series $\displaystyle\sum_{k\in\mathbb{N}}^{} a_{n_k}z^{n_k}$? Is $r$ related to $R$?
My first approach was thought that the radius was the same, but, it isn't true. Moreover, maybe the series doesn't converges for a certain values. Now I don't know how to proceed. Any hint?
The radius of convergence $R$ for the power series $\sum_{n=0}^{\infty} a_n z^n$ is given by $$ R=\frac1{\limsup_{n\rightarrow\infty} |a_n|^{\frac1n}}. $$ For the power series with a subsequence $a_{n_k}$, we have $$ r=\frac1{\limsup_{k\rightarrow\infty} |a_{n_k}|^{\frac1{n_k}}}. $$ So, we see that $R\leq r$.
Here is an example that the inequality can be strict.
$a_n = \begin{cases} 2^n \ \mbox{ if $n$ is even}\\ 2^{-n} \ \mbox{ if $n$ is odd}\end{cases}$.
Then $R=1/2$ but for the subsequence of all odd numbers, we have $r=2$.