I wanted to prove the convergence of the Riemann zeta function: \begin{align} \zeta(s)=\sum_{n=0}^\infty \frac{1}{n^s} \ \ \ \ \ \ \ \ \ \ \ \ \Re(s)>1 \end{align}
Here is my proof that uses the complex modulus.
Since \begin{align} \bigg\lvert \sum_i f(x_i) \bigg\rvert \leq \sum_i \lvert f(x_i) \rvert \end{align} Then \begin{align} \lvert \zeta(s) \rvert \leq \sum_{n=1}^\infty \bigg\lvert \frac{1}{n^s} \bigg\rvert = \sum_{n=1}^\infty \frac{1}{|n^\sigma||n^{it}|} = \sum_{n=1}^\infty \frac{1}{|n^\sigma||e^{it\ln(n)}|} = \sum_{n=1}^\infty \frac{1}{n^\sigma}, \end{align} for $\sigma=\Re(s) \in \mathbb{R}_+$\ $\{ 1 \}$ and $t=\Im(s) \in \mathbb{R}$.
Using the integral convergence-divergence test, $$\int_1^\infty \frac{1}{t^\sigma}dt = \frac{t^{1-\sigma}}{1-\sigma} \bigg\rvert_1^\infty = \lim_{t \to \infty} \frac{t^{1-\sigma}}{1-\sigma} - \frac{1}{1-\sigma}.$$ The integral is convergent for $\sigma=\Re(s)>1$, since then $1-\sigma$ is negative, and in turn $t^{1-\sigma}=1/t^k$, $k \in \mathbb{R}_+^*$, and $$as \ t \to \infty \ \ \ \ \ \ \ \ \ t^{1-\sigma} \to 0$$ Now, I think that this is rigorous to prove that $\lvert \zeta(s) \rvert$ is convergent for $\Re(s)>1$. Is this rigorous enough to prove that $\zeta(s)$ is convergent for $\Re(s)>1$?
In other words, since $\lvert \zeta(s) \rvert = \sqrt{\Re^2(\zeta(s))+\Im^2(\zeta(s))}$ and $\lvert \zeta(s) \rvert=conv.$, would there be needed any additional argument to conclude in a rigorous manner that the Riemann zeta function $\zeta(s)$ is convergent for $\Re(s)>1$?
In a more generalised form, for a function $f:\mathbb{C}\to \mathbb{C}$, if $|f(s)|$ is convergent for $s \in [a,b]$ (s.t. $[a,b] \in D=\mathbb{C}$) does that imply that $f$ is also convergent on the closed interval $[a,b]$?
Let $\{z_n\}$ denote a complex sequence, which without loss of generality doesn't contain $0$. Write $z_n=r_n(c_n+is_n)=x_n+iy_n$ with$$c_n,\,s_n,\,x_n,\,y_n\in\Bbb R,\,r_n:=|z_n|>0\implies|c_n|,\,|s_n|\in[0,\,1],\,c_n^2+s_n^2=1.$$If $\sum_nr_n$ is finite,$$X:=\Re\sum_nz_n=\sum_nr_nc_n,\,Y:=\Im\sum_nz_n=\sum_nr_ns_n$$both converge by this test, whence $\sum_nz_n=X+iY$ by the triangle inequality in $\Bbb C$.