It is well known the following function
$$ f(x) = \frac{1}{1^x} + \frac{1}{2^x} + \frac{1}{3^x} + \frac{1}{4^x} + \cdots $$
only converges if $x > 1$.
If we now consider $f(z)$ where z is complex, why can we say that the function converges for $Re(z) > 1$?
What's logic that allows us to simply apply convergence rules to the real part of a complex function's domain?
(I am not a trained mathematician so I'd appreciate answers which minimise assumptions about terminology.)
On
I am not an expert in this topic, but I think my proof is right.
Recall the triangle inequality for complex numbers $x,y$: $$|x+y|\le|x|+|y|$$
Note that $$\vert \frac1{n^z}\vert=\frac1{|e^{\ln(n)a}||e^{\ln(n)bi}|}=\frac1{n^a}$$
Thus, $$|\zeta(z)|=|\zeta(a+bi)|=|\sum_n\frac1{n^{a+bi}}|\le \sum_n\frac1{|n^{a+bi}|}=\sum_n \frac1{n^a} =\zeta(a)$$
Thus, if $\zeta(a)$ converges, $\zeta(a+bi)$ converges as well.
As indicated by Martin R in the comment, the reason is that absolute convergence of complex series implies convergence and in this case we have that for $z=x+iy$
$$\left|\frac1{n^z}\right|=\frac1{|n^z| }=\frac1{|n^{x}|}$$
indeed
$$|n^z|= |n^x||n^{iy}|=|n^x||e^{iy\log n}|=|n^x|$$
thus the series conveges for $Re(z)=x>1$.