Let $a,c_0,c_1,...\in \mathbb{R}$ with at least one of $c_0,c_1,c_2,...$ nonzero, and let the power series $\sum_{n=0}^{\infty}c_n(x-a)^n$ have positive radius of convergence $r$. Show that there exists a positive number $\delta<r$ such that the sum of the series is nonzero for every real number $x$ such that $0<|x-a|<\delta.$

Since it $r$ is its radius of convergence then we have that for $r>0$ the series converge for all $x$ such that $|x-a|<r$. But I am not sure how I can prove this.

How can I prove this?


1 Answer 1


Let $k:=\min \{n:c_n\ne 0\}$ and consider the expression

$$u(t)=t^k\left(|c_k|-\sum_{n=k+1}^\infty |c_n|t^{n+k}\right)$$

and its relation to values of the series for $|x-a|=t$.

Set $s(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$, then by the triangle inequality

$$|s(x)-c_k(x-a)^k|\le \sum_{n=k+1}^{\infty}|c_n|\,|x-a|^n=|c_k|\,|x-a|^k-u(|x-a|).$$

Again by the triangle inequality

$$|s(x)|\ge |c_k(x-a)^k|-|s(x)-c_k(x-a)^k|\ge u(|x-a|).$$

Now apply standard estimates to $u(t)$, for instance $$ u(\tfrac r2s)\ge s^k \left(|c_k|\left(\tfrac r2\right)^k-\sup_{n>k}|c_n|\left(\tfrac r2\right)^{n}\frac{s}{1-s} \right), $$ to find that it is positive for small positive arguments $t=\tfrac r2s$, $0\le s<1$, well inside the radius of convergence of the series, to find that $u(t)>0$ for small positive arguments $t$.

  • 1
    $\begingroup$ Can you elaborate please? $\endgroup$
    – user104235
    Feb 24, 2014 at 9:22

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