I am trying to parse out the below paragraph (click here and go to page 1), it is written in Definition 16.1. that-
It is easy to verify that this series converges absolutely and locally uniformly on $Re(s) > 1$ (use the integral test on an open ball strictly to the right of the line $Re(s) = 1$). By Theorem 16.17, it defines a holomorphic function on Re(s) > 1, since each term $e^{-s \log n}$ is holomorphic.
Theorem 16.17 states that-
Theorem 16.17. A sequence or series of holomorphic functions $f_n$ that converges locally uniformly on an open set $U$ converges to a holomorphic function $f$ on $U$, and the sequence or series of derivatives $f'_n$ then converges locally uniformly to $f'$ (and if none of the $f_n$ has a zero in $U$ and $f \neq 0$, then $f$ has no zeros in $U$).
What I understood is -
Point 1. Each term of $\zeta(s)$ is holomorphice since $n^{-s} = e^{-s \log n}$ and $e^{-s \log n}$ is holomorphic as $e^{-s \log n}$ is differentiable (we know $\frac{{{d}{\left({e}^{x}\right)}}}{{{\left.{d}{x}\right.}}}={e}^{x}$)
Point 2. $\zeta(s)$ converges (absolutely and locally uniformly) on $Re(s) > 1$ we can see this by the integral test (on an open ball strictly to the right of the line $Re(s) = 1$???)
Point 3. Finally we deduce that $\zeta(s)$ is holomorphic since it satisfies two conditions as written above (Point 1, 2) of Theorem 16.17.
Is this understanding is correct? Is the final result of the quoted paragraph is that $\zeta(s)$ is holomorphic ?
PS: it is not clear why the author wrote about "absolutely and locally uniformly" I learned about the integral test here, ( see Section 4-6 : Integral Test), there is no mention of absolutely or locally uniformly, can anyone comment on that (no answer is required)?