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 $|x| \leq 1$.