# Radius of convergence of a “subseries” of a series.

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?

## 3 Answers

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 $r=2$.

The radius of convergence $\sum a_nz^n$ is

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

Remember the $\limsup$ of a sequence is its greatest accumulation point. Your new sequence is like the old one, except a bunch of terms are replaced by $0$s. This might mean the greatest accumulation point of your new sequence is less than the old one, but of course it can't be greater.

Which in turn implies that your new radius is $\geq R$.

I'm not such an expert in Complex Analysis, but I guess you can't say much just for choosing indexes (finitely ou infinitely) without knowing the series very well.

Think for a moment in a series which terms even indexes are null. You can choose only the odd indexes and the radius of convergence doesn't change at all.

Think again for a moment in an alternate series. Deleting some indexes can make the series diverge with that radius of convergence.

Now, about an absolutely convergent series with only non-zero terms... that may be a good question.