# Does convergence of a complex series imply absolute convergence in certain cases?

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.

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.
• 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? – jesusbourne Mar 20 '17 at 3:04