# analytic continuation on boundary of a power series

I've having some confusion. I think this statement is true,

Suppose some power series is given say $P(z)$ in some neighborhood of $0$ whose radius of convergence is $R$ and then prove $P$ can't be analytic on boundary of $B(0,R)$

What I did was like, suppose $P$ is analytic on whole boundary, then for each point we'll get a open ball where $P$ is analytic but as boundary is compact so we can choose finitely many points which covers whole boundary so we can get slightly bigger $r>R$ such that $P$ is analytic in $B(0,r)$. But then integrating over some circle of radius $R<s<r$ wee see $n$th term of power series $|a_n|\leq \frac{M(s)}{s^n}$ and so radius of convergence is at least $s\implies$ greater than $R$. Am I wrong ?

• You are right, and your approach is correct. – Mario Carneiro Feb 21 '15 at 7:30
• yesh, so my proof is correct right ? our prof mentioned the statement but told to think.. – dragoboy Feb 21 '15 at 7:32