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Let $\sum_{n=0}^\infty a_n x^n$ be a power series. If the series converges for some $x_0$ then the series will converge absolutely for all $|x|<|x_0|$.

The proof of the statement is straightforward: its terms converge to zero, so they are bounded $|a_nx_{0}^n|\leq M$ and for all $|x|<|x_0|$ the series converges absolutely because $$ |a_nx^n|=|a_nx_0^n|\Big|\frac{x}{x_0}\Big|^n \leq Mr^n, \quad r = \Big|\frac{x}{x_0}\Big|<1 $$

Is the proof the same for complex-valued series or there are some notable differences that we should pay attention to?

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It is the same, provided you know the following lemma:

Let $(V,\lVert\bullet\rVert)$ be a normed real vector space. The following are equivalent:

  1. $V$ is a Banach space

  2. For any sequence $(v_n\,:\,n\in\Bbb N)$ such that $\sum\limits_{n=0}^\infty \lVert v_n\rVert<\infty$ there is $s\in V$ such that $\lim\limits_{m\to\infty} \left\lVert s-\sum\limits_{n=0}^m v_n\right\rVert=0$

(but this seems the case here, since you already know that absoultely convergent complex series are convergent)

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  • $\begingroup$ Thank you for your reply! Yes, I concur that absolute convergence in a Banach space implies convergence. So the proof is completely the same and in the proof all moduli of complex numbers will become just real numbers? $\endgroup$ – Konstantin Feb 9 '17 at 19:16
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    $\begingroup$ The moduli of complex numbers are real numbers. If memory serves me right, the proof is written the same way, with $\lvert x\rvert$ standing for $(\overline xx)^{1/2}$. $\endgroup$ – user228113 Feb 9 '17 at 20:36

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