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!


1 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 $\sum_{i=1}^\infty b_n z^n$?

  • $\begingroup$ Yes I edited the question thanks for pointing that out $\endgroup$
    – Heisenberg
    Nov 30, 2014 at 6:00
  • $\begingroup$ formalizing that is the issue but since $b_n$ terms are a subset of the $a_n$ terms I kinda intuitively get it $\endgroup$
    – Heisenberg
    Nov 30, 2014 at 6:01
  • $\begingroup$ Can you show that if $A \subset B$, $\sup A \le \sup B$? $\endgroup$
    – Peter R
    Nov 30, 2014 at 20:32

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