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! :)


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! :) $\endgroup$ – M.Gonzalez Dec 31 '18 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.