Is this a valid proof of absolute convergence of $ \zeta(s) $ for $ \Re(s) > 1 $ and divergence of $ \zeta(s) $ for $ \Re(s) < 0 $?
By integral test for convergence, $ \sum_{n=1}^{\infty} \left| \frac{1}{n^s} \right| $ converges if and only if $ \int_1^{\infty} \left| \frac{1}{x^s} \right| dx $ converges.
Let $ \sigma = \Re(s) $.
For $ \sigma = 1 $,
$$ \int_1^{\infty} \left| \frac{1}{x^s} \right| dx = \int_1^{\infty} \frac{1}{x} dx = \left[ \log x \right]_{1}^{\infty} = \infty. $$
For $ \sigma \ne 1 $,
$$ \int_1^{\infty} \left| \frac{1}{x^s} \right| dx = \int_1^{\infty} \frac{1}{x^{\sigma}} dx = \frac{1}{(1 - \sigma)} \left[\frac{1}{x^{\sigma - 1}}\right]_{1}^{\infty} = \begin{cases} \infty & \text{when} & \sigma < 1, \\ \frac{1}{\sigma - 1} & \text{when} & \sigma > 1. \end{cases} $$
This shows that $ \int_1^{\infty} \left| \frac{1}{x^s} \right| dx $ converges for $ \Re(s) > 1 $ and diverges for $ \Re(s) \le 1 $.
Therefore by integral test of convergence $ \sum_{n=1}^{\infty} \left| \frac{1}{n^s} \right| $ converges for $ \Re(s) > 1 $ and diverges for $ \Re(s) \le 1 $.
