# Comparison Test for complex series: logical argument?

I am trying to show that the comparison test holds for complex series, meaning: if $$\sum_{n=0}^{\infty} z_n$$ is a complex series and $$\sum_{n=0}^{\infty} a_n$$ is a convergent non-negative real number series and $$|z_n| \leq a_n \ \forall n \in \mathbb{N}$$ then $$\sum_{n=0}^{\infty} z_n$$ converges.

To do so, I wanted to show that $$|z_n - z_m| \leq |a_n - a| + |a_m - a| < \epsilon$$ to get a Cauchy convergent series. My question is, is it logical my argument as it follows?

We have:

$$|z_n - z_m|^2 = |z_n|^2 + |z_m|^2 - (z_n\overline{z_m}) - (\overline{z_n}z_m) = |z_n|^2 + |z_m|^2 - 2Re(\overline{z_n}z_m) \leq (|z_n| - |z_m|)^2$$ because $$Re(z) \leq |z|$$. Hence:

$$|z_n - z_m| \leq |z_n| - |z_m|$$

and since $$|z_n| \geq 0, \ \ \forall n \in \mathbb{N}$$, it is obvious that $$|z_n| - |z_m| \leq |z_n|+|z_m|$$.

Is my argument logical or am I missing a point?

Thank you people so much for your help! :)

• Which number is $a$? Dec 30, 2018 at 17:50

First, you want to show that $$\sum_{n=0}^\infty z_n$$ converges, but you check whether $$(z_n)_n$$ is a Cauchy sequence, which is something different (If $$z_n=1$$ for all $$n$$ then $$(z_n)_n$$ is a Cauchy sequence but the series is divergent).
Second, you have the inequality $$|z_n|^2 + |z_m|^2 - 2Re(\overline{z_n}z_m) \stackrel{??}{\leq} (|z_n| - |z_m|)^2$$ backwards (the LHS is greater than or equal to the RHS), since from $$\operatorname{Re} z \leq |z|$$ you get $$-\operatorname{Re} z \geq -|z|$$.
I would do this proof by looking at the partial sums $$Z_N = \sum_{n=0}^N z_n$$ and $$A_N = \sum_{n=0}^N a_n$$ and use the fact that $$(A_N)_N$$ is a Cauchy sequence to prove that $$(Z_N)_N$$ is a Cauchy sequence.