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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.

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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.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Alfdav Nov 14 '18 at 21:17

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