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! :)

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

1 Answer 1


I think you are confusing a few things:

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.

  • $\begingroup$ Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :) $\endgroup$ Dec 31, 2018 at 0:46

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