[proofplan]
We reduce absolute convergence of the complex Dirichlet series $\sum n^{-s}$ to convergence of the real series $\sum n^{-\sigma}$, where $\sigma = \operatorname{Re}(s)$. This reduction rests on the elementary identity $|n^{-s}| = n^{-\sigma}$, which uses only the principal branch of the complex exponential and the fact that $|e^{it \log n}| = 1$ for $t \in \mathbb{R}$. The real $p$-series $\sum n^{-\sigma}$ is known to converge if and only if $\sigma > 1$ by the integral test, so absolute convergence of the Dirichlet series holds iff $\operatorname{Re}(s) > 1$.
[/proofplan]
[step:Define the series and decompose the complex variable]
For $s \in \mathbb{C}$, consider the Dirichlet series
\begin{align*}
\zeta_{\mathrm{ser}}: \{s \in \mathbb{C} : \text{series converges absolutely}\} &\to \mathbb{C} \\
s &\mapsto \sum_{n = 1}^{\infty} n^{-s},
\end{align*}
where $n^{-s} := e^{-s \log n}$ using the real natural logarithm (valid since $n \geq 1$, so $\log n \geq 0$ is real). Write $s = \sigma + i t$ with $\sigma = \operatorname{Re}(s) \in \mathbb{R}$ and $t = \operatorname{Im}(s) \in \mathbb{R}$.
The series $\sum_{n \geq 1} n^{-s}$ converges absolutely if and only if $\sum_{n \geq 1} |n^{-s}|$ converges in the ordinary sense of a non-negative real series. Our goal is to compute $|n^{-s}|$ and reduce to a known real $p$-series.
[/step]
[step:Reduce $|n^{-s}|$ to the real expression $n^{-\sigma}$]
Using the definition $n^{-s} = e^{-s \log n}$ and the factorisation $s = \sigma + it$,
\begin{align*}
n^{-s} = e^{-(\sigma + it) \log n} = e^{-\sigma \log n} \cdot e^{-it \log n} = n^{-\sigma} \cdot e^{-it \log n},
\end{align*}
where $n^{-\sigma} := e^{-\sigma \log n}$ is the real positive number defined by the real exponential. Taking moduli and using multiplicativity of the complex absolute value,
\begin{align*}
|n^{-s}| = |n^{-\sigma}| \cdot |e^{-it \log n}| = n^{-\sigma} \cdot 1 = n^{-\sigma},
\end{align*}
since $n^{-\sigma} > 0$ is real positive (so its modulus is itself) and $|e^{i \theta}| = 1$ for every $\theta \in \mathbb{R}$ (applied with $\theta = -t \log n \in \mathbb{R}$).
Therefore
\begin{align*}
\sum_{n = 1}^{\infty} |n^{-s}| = \sum_{n = 1}^{\infty} n^{-\sigma}.
\end{align*}
The series on the left converges iff the series on the right does.
[/step]
[step:Apply the integral test for the real $p$-series $\sum n^{-\sigma}$]
We prove the classical fact: $\sum_{n \geq 1} n^{-\sigma}$ converges iff $\sigma > 1$.
(a) Assume $\sigma > 1$. The function
\begin{align*}
f: [1, \infty) &\to (0, \infty), & t &\mapsto t^{-\sigma}
\end{align*}
is positive, continuous, and strictly decreasing (its derivative $-\sigma t^{-\sigma - 1}$ is negative). By the [integral test](/theorems/???) for series with positive decreasing terms, $\sum_{n \geq 1} f(n)$ and $\int_1^{\infty} f(t) \, d\mathcal{L}^1(t)$ converge or diverge together. We compute
\begin{align*}
\int_1^{\infty} t^{-\sigma} \, d\mathcal{L}^1(t) = \lim_{R \to \infty} \int_1^R t^{-\sigma} \, d\mathcal{L}^1(t) = \lim_{R \to \infty} \frac{R^{1 - \sigma} - 1}{1 - \sigma} = \frac{1}{\sigma - 1},
\end{align*}
where the limit $R^{1 - \sigma} \to 0$ uses $1 - \sigma < 0$. The integral converges, hence so does the series.
(b) Assume $\sigma \leq 1$. If $\sigma \leq 0$, then $n^{-\sigma} = n^{|\sigma|} \geq 1$ for all $n \geq 1$, so $\sum n^{-\sigma}$ diverges by comparison with $\sum 1$. If $0 < \sigma \leq 1$, the same integral test applies: $\int_1^R t^{-\sigma} \, d\mathcal{L}^1(t)$ equals $(R^{1-\sigma} - 1)/(1 - \sigma)$ for $\sigma < 1$ (diverging as $R \to \infty$ since $1 - \sigma > 0$) and equals $\log R \to \infty$ for $\sigma = 1$. In both sub-cases the integral diverges, so the series diverges.
[/step]
[step:Combine the reduction and the $p$-series dichotomy]
By Step 2, the Dirichlet series $\sum n^{-s}$ converges absolutely iff $\sum n^{-\sigma}$ converges. By Step 3, $\sum n^{-\sigma}$ converges iff $\sigma > 1$. Since $\sigma = \operatorname{Re}(s)$, absolute convergence of $\sum n^{-s}$ is equivalent to $\operatorname{Re}(s) > 1$, which proves the theorem.
[/step]
