# Complex power series - Radius of convergence

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!

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 $$\sum_{i=1}^\infty b_n z^n$$?
• formalizing that is the issue but since $b_n$ terms are a subset of the $a_n$ terms I kinda intuitively get it Nov 30, 2014 at 6:01
• Can you show that if $A \subset B$, $\sup A \le \sup B$? Nov 30, 2014 at 20:32