Theorem 1. Given the series

(1)

Proof. Let $\varepsilon$ be an arbitrary positive number. Denote the partial sum of (1) by $$S_n = z_1+\ldots+z_n = (a_1+ib_1)+\ldots+(a_n+ib_n) = (a_1+\ldots+a_n)+i(b_1+\ldots+b_n) := A_n+iB_n$$ ($n = 1,\,2,\,3,\,\ldots$ ). When (1) converges to the sum $A+iB$ , then there is a number $n_\varepsilon$ such that for any integer $n > n_\varepsilon$ we have $$|(A_n-A)+i(B_n-B)| = |(A_n+iB_n)-(A+iB)| < \varepsilon.$$ But a complex number is always absolutely at least equal to the real part (see the inequalities in modulus of complex number), and therefore $|A_n-A| \leqq |(A_n-A)+i(B_n-B)| < \varepsilon$ , similarly $|B_n-B| \leqq |(A_n-A)+i(B_n-B)| < \varepsilon$ as soon as $n > n_\varepsilon$ . Hence, $A_n \to A$ and $B_n \to B$ as $n \to \infty$ . This means the convergences $$a_1+a_2+a_3+\ldots = A\;\;\;\mbox{and}\;\;\;b_1+b_2+b_3+\ldots = B,$$ Q.E.D. The converse part is straightforward.

Theorem 2. Notations same as in the preceding theorem. The series $$|z_1|+|z_2|+|z_3|+\ldots$$ converges if and only if the series $$a_1+a_2+a_3+\ldots\;\;\;\mbox{and}\;\;\;b_1+b_2+b_3+\ldots$$ converge absolutely.

Proof. Use the inequalities $$0 \leqq |a_n| \leqq |z_n|,\quad 0 \leqq |b_n| \leqq |z_n|$$ and $$0 \leqq |z_n| \leqq |a_n|+|b_n|$$ for using the comparison test.

Theorem 3. If the series $\displaystyle\sum_{n=1}^\infty|z_n|$ converges, then also the series $\displaystyle\sum_{n=1}^\infty z_n$ converges and we have $$\left|\sum_{n=1}^\infty z_n\right| \leqq \sum_{n=1}^\infty|z_n|.$$

Proof. By theorem 2, the convergence of $\sum|z_n|$ implies the convergence of $\sum a_n$ and $\sum b_n$ , which, by theorem 1, in turn imply the convergence of $\sum z_n$ . Since for every $n$ the triangle inequality guarantees the inequality $$\left|\sum_{j=1}^n z_j\right| \leqq \sum_{j=1}^n|z_j|,$$ then we must have the asserted limit inequality, too.