# Showing that a power series is bounded

Let $$\sum a_nx^n$$ be a power series with a radius of convergence of $$2$$. Then there exists an $$M$$ such that

$$\left\lvert \sum_{n=1}^\infty a_nx^n\right\rvert \le M\left\lvert x \right\rvert$$

where $$\left\lvert x \right\rvert \le 1$$

The constant term is $$0$$.

I'm not sure where to exactly to start. I've tried showing the series is bounded and attempted to rearrange the expression into the given form, but that was not successful.

$$g(x)=\sum\limits_{n=1}^{\infty}a_nx^{n-1}$$ also has radius of convergence $$2$$. Hence $$g$$ is analytic, in particular continuous, for $$|x| \leq 1$$. Since continuous functions are bounded on compact sets there exists $$M <\infty$$ such that $$|g(x)| \leq M$$ for $$|x| \leq 1$$. Hence $$|f(x)|=|xg(x)| \leq M|x|$$ for $$|x| \leq 1$$.