[proofplan]
We bound each term of the Dirichlet $L$-series by the corresponding term of a real $p$-series. For pointwise absolute convergence, the exponent is $\sigma = \operatorname{Re}(s) > 1$. For [uniform convergence](/page/Uniform%20Convergence) on a compact set, compactness gives a uniform margin $\operatorname{Re}(s) \geq 1+\delta$, so the Weierstrass test applies with the summable majorant $n^{-1-\delta}$. Finally, each summand is holomorphic in $s$, and local uniform convergence of holomorphic functions preserves holomorphy.
[/proofplan]
[step:Bound the absolute value of each summand by a real $p$-series term]
Let $H := \{s \in \mathbb{C} : \operatorname{Re}(s) > 1\}$. For each $n \in \mathbb{N}$, define
\begin{align*}
f_n: H &\to \mathbb{C} \\
s &\mapsto \frac{\chi(n)}{n^s}.
\end{align*}
Since $\chi$ is a Dirichlet character, $|\chi(n)| \leq 1$ for every $n \in \mathbb{N}$. If $s = \sigma + it \in H$, with $\sigma := \operatorname{Re}(s)$ and $t := \operatorname{Im}(s)$, then
\begin{align*}
|n^{-s}|
= |\exp(-s\log n)|
= \exp(-\sigma \log n)
= n^{-\sigma}.
\end{align*}
Therefore
\begin{align*}
\left|\frac{\chi(n)}{n^s}\right|
= |\chi(n)|\,|n^{-s}|
\leq n^{-\sigma}.
\end{align*}
[/step]
[step:Deduce absolute convergence at every point of the half-plane]
Fix $s \in H$, and write $\sigma := \operatorname{Re}(s)$. Then $\sigma > 1$. By the estimate from the previous step,
\begin{align*}
\sum_{n=1}^{\infty} \left|\frac{\chi(n)}{n^s}\right|
\leq
\sum_{n=1}^{\infty} n^{-\sigma}.
\end{align*}
The real $p$-series on the right converges because $\sigma > 1$. Hence the Dirichlet $L$-series
\begin{align*}
\sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}
\end{align*}
converges absolutely for every $s \in H$.
[/step]
[step:Use compactness to obtain a uniform summable majorant]
Let $K \subset H$ be compact. Define
\begin{align*}
\rho: K &\to \mathbb{R} \\
s &\mapsto \operatorname{Re}(s).
\end{align*}
The function $\rho$ is continuous, and $K$ is compact, so $\rho$ attains its minimum on $K$. Let
\begin{align*}
m := \min_{s \in K} \operatorname{Re}(s).
\end{align*}
Since $K \subset H$, we have $m > 1$. Define $\delta := m - 1 > 0$. Then for every $s \in K$ and every $n \in \mathbb{N}$,
\begin{align*}
\left|\frac{\chi(n)}{n^s}\right|
\leq n^{-\operatorname{Re}(s)}
\leq n^{-1-\delta}.
\end{align*}
The majorant series
\begin{align*}
\sum_{n=1}^{\infty} n^{-1-\delta}
\end{align*}
converges because $1+\delta > 1$. By the Weierstrass uniform convergence test, the series
\begin{align*}
\sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}
\end{align*}
converges uniformly on $K$.
[guided]
We now strengthen pointwise convergence to uniform convergence on compact subsets. The important point is that on a compact set inside the open half-plane $\operatorname{Re}(s) > 1$, the real parts of the points cannot approach the boundary line $\operatorname{Re}(s)=1$.
Let $K \subset H$ be compact. Define the real-valued function
\begin{align*}
\rho: K &\to \mathbb{R} \\
s &\mapsto \operatorname{Re}(s).
\end{align*}
This function is continuous because the real-part map is continuous on $\mathbb{C}$. Since $K$ is compact, $\rho$ attains a minimum:
\begin{align*}
m := \min_{s \in K} \operatorname{Re}(s).
\end{align*}
Every point of $K$ lies in $H$, so every point of $K$ has real part strictly larger than $1$. Because the minimum is attained at a point of $K$, this gives $m > 1$. Set $\delta := m - 1$, so $\delta > 0$ and
\begin{align*}
\operatorname{Re}(s) \geq 1+\delta
\end{align*}
for every $s \in K$.
Using the termwise bound already proved, for every $s \in K$ and every $n \in \mathbb{N}$ we obtain
\begin{align*}
\left|\frac{\chi(n)}{n^s}\right|
\leq n^{-\operatorname{Re}(s)}
\leq n^{-1-\delta}.
\end{align*}
The right-hand side no longer depends on $s$. This is exactly the kind of estimate required by the Weierstrass uniform convergence test: the numerical series
\begin{align*}
\sum_{n=1}^{\infty} n^{-1-\delta}
\end{align*}
converges because $1+\delta > 1$, so the series
\begin{align*}
\sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}
\end{align*}
converges uniformly on $K$.
[/guided]
[/step]
[step:Conclude holomorphy from locally uniform convergence of holomorphic summands]
For each $n \in \mathbb{N}$, the function $f_n: H \to \mathbb{C}$, $s \mapsto \chi(n)\exp(-s\log n)$, is holomorphic on $H$ because it is a constant multiple of the exponential of an affine function of $s$. The previous step proves that $\sum_{n=1}^{\infty} f_n$ converges uniformly on every compact subset of $H$. By the theorem that a locally uniform limit of holomorphic functions is holomorphic (citing a result not yet in the wiki: Weierstrass theorem on locally uniform limits of holomorphic functions), the sum
\begin{align*}
L(\cdot,\chi): H &\to \mathbb{C} \\
s &\mapsto \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}
\end{align*}
is holomorphic on $H$. This proves the stated absolute convergence, compact-uniform convergence, and holomorphy.
[/step]
