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Suppose that the complex series $$ \sum_{n=0}^{\infty}c_n\xi^n $$ converges for some complex number $\xi\neq 0$. Show that the complex series $$ \sum_{n=0}^{\infty}c_nw^n $$ converges absolutely if $|w|<|\xi|$ without using the definition of radius of convergence. My problem with this question is the absolute values. I can't say $$ \Bigg|\sum_{n=0}^{\infty}c_n\xi^n\Bigg|=\sum_{n=0}^{\infty}|c_n\xi^n| $$ because convergence does not imply absolute convergence. My line of thinking is that I have to state with justification that $\Bigg|\sum_{n=0}^{\infty}c_n\xi^n\Bigg|=\sum_{n=0}^{\infty}|c_n\xi^n|$ and that $\sum_{n=0}^{\infty}|c_n\xi^n|$ converges. From there I think I can apply the relation $|w|<|\xi|$ to the series and prove the statement. How do I work around the issue of convergence not implying absolute convergence without knowing anything about the first series? I'm imagining cases where my first series is a harmonic series and absolute convergence doesn't follow from convergence.

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Hint: You actually only need that $\left|c_n\xi^n\right|$ is bounded, which is much weaker than needing that $\sum c_n\xi^n$ converges.

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  • $\begingroup$ I'm going to rewrite the Comparison Test from Abbott's Understanding Analysis, $2^{\text{nd}}$ edition: Assume $(c_n\xi^n)$ and $(c_nw^n)$ are sequences satisfying $0\leq |c_nw^n|\leq |c_n\xi^n|$ for all $n\in\mathbb{N}$. If $\sum_{n=1}^{\infty}c_n\xi^n$ converges, then $\sum_{n=1}^{\infty}c_nw^n$ converges. Is it correct for me to just extend this real analysis theorem to complex analysis? $\endgroup$ – jesusbourne Mar 20 '17 at 3:04
  • $\begingroup$ I am not familiar with this book. Examine the proof in Abbott. Does any step in the proof not hold if any of the values are not real? $\endgroup$ – DanielWainfleet Mar 20 '17 at 8:52

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