Why this proof of 'absolute convergence $\implies$ convergence' works for complex sequences? I'm reading this wikipedia proof about the absolute convergence of a complex sequence implying convergence, but it considers a real sequence. Why does it works for complex ones?
 A: I feel like I owe you a proper answer, because I have neglected to read the proof in question before commenting, so here it goes.
I'm assuming that you agree with the proof of the real case. So let me assume that if a series of real numbers converges absolutely, then it also converges in the traditional sense.  
Take the complex series $\{s_n\}_{n\in \mathbb{N}} = \{\sum_{j=0}^{n}(a_j + ib_j)\}_{n\in \mathbb{N}}$ where $a_j,b_j \in \mathbb{R}$. Assume this series converges absolutely, that is $\sum_{j=0}^{\infty}|a_j + ib_j| < \infty$. We want to show that this series converges in the traditional sense.
Since $|a_j| \leq |a_j+ib_j|$ and $|b_j| = |ib_j| \leq |a_j+ib_j|$, we have that the real series $\{\sum_{j=0}^{n}a_j\}_{n\in \mathbb{N}}$ and $\{\sum_{j=0}^{n}b_j\}_{n\in \mathbb{N}}$ are both absolutely convergent, so by our assumption, they are also convergent.
Notice that our original complex series $\{s_n\}_{n\in \mathbb{N}}$ is also a complex sequence and its real part is $\{\sum_{j=0}^{n}a_j\}_{n\in \mathbb{N}}$, while its imaginary part is $\{\sum_{j=0}^{n}b_j\}_{n\in \mathbb{N}}$. 
To conclude, we've shown that $\{s_n\}_{n\in \mathbb{N}}$ is a complex sequence with convergent real and imaginary parts thus it's convergent in the complex sense.
Thus absolute convergence implies convergence in the case of complex series, which is no surprise, since the complex numbers form a Banach space with the usual complex absolute value, and in a Banach space, every absolutely convergent series is also convergent.
