Absolutely convergent series of complex functions.

I have to do the following excercise:

Let $\{f_n(z)\}_{n\in\mathbb{N}}$ a sequence of complex functions, and let $\sum_{n=1}^\infty f_n(z)$.

Prove that: if $\sum_{n=1}^\infty |f_n(z)|$ converges, then $\sum_{n=1}^\infty f_n(z)$ converges.

I know how to prove it for a series $\sum_{n=1}^\infty z_n$ of complex numbers with $z_n=x_n+iy_n$ because if $\sum_{n=1}^\infty |z_n|$ converges, one can observe that $|x_n|<|z_n|$ and $|y_n|<|z_n|$ then by the comparison criteria the real numbers series $\sum_{n=1}^\infty |x_n|$ and $\sum_{n=1}^\infty |y_n|$ converge and we know for real series that this implies that $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ converge.

If we call $R_n=\sum_{k=1}^n x_n$, $I_n=\sum_{k=1}^n y_n$ and $S_n=\sum_{k=1}^n z_n$.

And $\lim_{n \rightarrow \infty}R_n=x$, $\lim_{n \rightarrow \infty}I_n=y$, then

$$\lim_{n \rightarrow \infty}S_n=\lim_{n \rightarrow \infty}R_n+i\lim_{n \rightarrow \infty}I_n=x+iy.$$

Then $S_n$ converges and $\sum_{n=1}^\infty z_n$ does as well.

Is it enough to call $\{w_n\}=\{f_n(z)\}$ in my original problem and just apply this proof?

Thanks in advance.

Answer

Yes, that would be correct. On the other hand, you don't have to decompose your series into real and imaginary part. Suppose that $\sum_{n=1}^\infty\lvert z_n\rvert$ converges. Take $\varepsilon>0$. Then there is a natural $N$ such than $$m\geqslant n\geqslant N\implies \sum_{k=n}^m\lvert z_k\rvert<\varepsilon,$$ and therefore, by the triangle inequality, $$m\geqslant n\geqslant N\implies\left\lvert\sum_{k=n}^mz_k\right\rvert<\varepsilon.$$ Therefore, by Cauchy's criterion, the series $\sum_{n=1}^\infty z_n$ converges too.