Complex power series - Radius of convergence

Question

Let $f$ be analytic in the unit disk. Then we can write that as, $$f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n,|z|<1$$ Now let $a_n=\frac{f^{(n)}(0)}{n!}$. So the radius of convergence $R$ is $$R=\frac{1}{\limsup|a_n|^{1/n}}$$

Say I define $b_n=a_{2n}$. What can be said about the analiticity and radius of convergence of the power series $\sum_{n=0}^{\infty}b_{n}z^n$? I feel like it should also definitely be analytic in the unit disk but don't know how to gie a proper proof. Hope someone can help out thanks!

Answer 1

I'm assuming that you're intending to ask about the series $\sum_{i=1}^\infty b_n z^n$. First, note that it is trivially true that any function defined by a power series is analytic anywhere in its radius of convergence. Now, note that $\limsup|a_n| \ge \limsup|b_n|$. (Why? Can you formalize this?). Then what does that tell you about the radius of convergence of $\sum_{i=1}^\infty b_n z^n$?