It is well known that the Riemann zeta function $\zeta(s)$ over the complex domain converges for $\sigma>1$, where $s=\sigma+it$.
The most common proofs use the triangle inequality to show that that
$$\left|\sum\frac{1}{n^{s}}\right|\leq\sum\left|\frac{1}{n^{\sigma}}\frac{1}{n^{it}}\right|$$
and use $n^{it} = e^{it\ln(n)}$ to make clear $n^{it}$ has a magnitude of 1. This leaves us with
$$\left|\sum\frac{1}{n^{s}}\right|\leq\sum\frac{1}{n^{\sigma}}$$
We know from other methods (comparing sums to integrals, for example) that $\sum\frac{1}{n^{\sigma}}$ converges for $\sigma>1$.
Those proofs then conclude that the complex zeta function $\zeta(s)$ converges for $s>1$ because the magnitude of the sum is dominated by the real valued $\zeta(\sigma)$.
Question: The above logic seems to exclude showing that there is no $\sigma<1$ for which the complex $\zeta(s)$ converges.
What have I misunderstood?
As context, the integral test for real $\zeta(\sigma)$ for does test for both possibilities (link).