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Let $(z_n)$ be a sequence of complex numbers. If $\sum_{n=1}^\infty \vert z_n\vert$ converges then $\sum_{n=1}^\infty z_n$ does too.

Show that this statement is true assuming we know this is true for real series.


Let $z_n=a_n+ib_n$ where $(a_n)$ and $(b_n)$ are real sequences. Then

\begin{align} \sum_{n=1}^\infty \vert z_n\vert &= \sum_{n=1}^\infty \vert a_n + ib_n \vert \\ &\ge \sum_{n=1}^\infty \Big[\vert a_n \vert - \vert b_n \vert\Big] \\ &= \sum_{n=1}^\infty \vert a_n \vert - \sum_{n=1}^\infty\vert b_n \vert \end{align}

So if the infinite series of $\vert z_n\vert$ converges then $\vert a_n\vert$ and $\vert b_n\vert$ converge. Now also note that

$$\sum_{n=1}^\infty z_n = \sum_{n=1}^\infty a_n + i\sum_{n=1}^\infty b_n.$$

Since the infinite series of $|a_n|$ and $|b_n|$ converge, then the infinite series of $a_n$ and $b_n$ converge. Hence

$$\sum_{n=1}^\infty z_n$$

converges.


Is this a sufficient proof?

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No, it's not. Having three sequences, $r_n,s_n,t_n$ where $r_n$ converges, we cannot conclude from $r_n\geq s_n-t_n$ that $s_n$ and $t_n$ converge. For example, let $r_n=2$, $s_n=n+1$ and $t_n=n$. See what happens there?


But we can prove it by doing the following. \begin{align} \sum_{n=1}^{\infty}|z_n|&=\sum_{n=1}^{\infty}|a_n+ib_n|\\ &= \sum_{n=1}^{\infty}\sqrt{a_n^2+b_n^2}\\ &\geq\sum_{n=1}^{\infty}\sqrt{a_n^2+0}\\ &=\sum_{n=1}^{\infty}|a_n| \end{align} We can do the same to show $|b_n|$ converges, and then we can proceed to reason the way you did, for the last part of your proof was correct.

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