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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.

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$g(x)=\sum\limits_{n=1}^{\infty}a_nx^{n-1}$ 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 $|x| \leq 1$.

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  • $\begingroup$ I see, thank you! $\endgroup$ – J.Doe Feb 2 '19 at 12:56

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