how to prove this statement related to radius of convergence

Suppose that the power series $$\sum b_nx^n$$ converges for $|x|$ less than or equal to $1$.

Suppose that for some $s$ greater than $0$, $p(x)=0$ for all $|x|$ less than $s$.

How to show that $b_n=0$ for all $n$ greater than equal to $1$?

• Is $p(x)=\sum b_n x^n$? Feb 10 '16 at 3:12
• Do you know about the uniqueness property of analytic functions ? Feb 10 '16 at 3:17
• yes i do know,.. Feb 10 '16 at 3:28

Based on the context of learning, if the textbook said you can differentiate a power series term by term inside circle of the convergence, then just differentiate it. You can prove $c_1 =0$ and do it recursively.
• Which you can show easily and indirectly by showing that you can integrate $\sum _{n>0}n a_n x^{n-1}$ term by term, because the sequence for the derivative converges uniformly on any closed set inside the open circle of convergence. Feb 10 '16 at 5:59
You don't say what $n$ runs over in the sum. Let's say from $1$ to $\infty$.
Factor out $x$. You thus have $x\sum_{n=1}^\infty b_n x^{n-1}$. The latter series has the same radius of convergence as the former by the formula in Baby Rudin 3.39. Divide by $x$. You thus have $\sum_{n=1}^\infty b_n x^{n-1}=0$ for $|x|<s$. Let $x=0$. You thus have $b_1=0$.
Suppose you've proved $b_1, b_2, \ldots, b_m=0$. We want to show $b_{m+1}=0$. We have $p(x)=\sum_{n=m+1}^\infty b_n x^n=0$ for $|x|<s$. Factor out $x$ again, divide by $x$, and let $x=0$. We get $b_{m+1}=0$ as desired.