convergence of complex term series

A series

${\displaystyle \sum _{\nu =1}^{\mathrm{\infty }}}{c}_{\nu }=c_1+c_2+c_3+\mathrm{\dots }$(1)

with complex terms

$$c_{\nu }=a_{\nu }+i{b}_{\nu }\mathit{\hspace{1em}\hspace{1em}}(a_{\nu },b_{\nu }\in ℝ\forall \nu )$$

is convergent iff the sequence of its partial sums converges to a complex number.

Theorem 1.  The series (1) converges iff the series

${\displaystyle \sum _{\nu =1}^{\mathrm{\infty }}}{a}_{\nu }\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\displaystyle \sum _{\nu =1}^{\mathrm{\infty }}}{b}_{\nu }$(2)

formed by real parts and the imaginary parts of its terms both are convergent.

Proof.  Let  $\epsilon >0$.  Denote

$$\sum _{\nu =1}^n{a}_{\nu }:=s_n,\sum _{\nu =1}^n{b}_{\nu }:=t_n,\sum _{\nu =1}^n{c}_{\nu }:=u_n.$$

If the series (2) are convergent with sums $S$ and $T$, then there is a number $N$ such that

Accordingly,

i.e. the series (1) converges to $S+iT$.  If, conversely, (1) converges to a complex number

$$u=s+it\mathit{\hspace{1em}}(s,t\in ℝ),$$

then

$$|s_n-s|\leqq |(s_n-s)+i(t_n-t)|=|u_n-u|,|t_n-t|\leqq |(s_n-s)+i(t_n-t)|=|u_n-u|,$$

and consequently,  $\underset{n\to \mathrm{\infty }}{lim}{s}_n=s$  and  $\underset{n\to \mathrm{\infty }}{lim}{t}_n=t$, i.e. the series (2) are convergent with sums the real numbers $s$ and $t$.

Theorem 2.  The series (1) converges absolutely iff the series (2) both converge absolutely.

Proof.  The absolute convergence of (1) means that the series

$$\sum _{\nu =1}^{\mathrm{\infty }}|c_{\nu }|$$

converges.  But since  ${|c_{\nu }|}^2={|a_{\nu }|}^2+{|b_{\nu }|}^2$,  we have

$$|a_{\nu }|\leqq |c_{\nu }|;|b_{\nu }|\leqq |c_{\nu }|\leqq |a_{\nu }|+|c_{\nu }|.$$

From these inequalities we can infer the assertion of the theorem 2.