[proofplan] The proof converts absolute convergence at one point into a summable numerical majorant for the coefficients. On the shifted half-plane $\operatorname{Re}(s) \geq \sigma_0 + \varepsilon$, each term $|a_n n^{-s}|$ is bounded by the corresponding summable term $|a_n|n^{-\sigma_0}$. The uniform Cauchy criterion for numerical tails then gives [uniform convergence](/page/Uniform%20Convergence) of the series of absolute values, which is exactly uniform absolute convergence of the original Dirichlet series. [/proofplan] [step:Rewrite absolute convergence at $s_0$ as a summability condition] For each $n \in \mathbb{N}$, the definition $n^{-s_0} = e^{-s_0 \log n}$ gives \begin{align*} |n^{-s_0}| = |e^{-s_0 \log n}| = e^{-\operatorname{Re}(s_0)\log n} = n^{-\sigma_0}. \end{align*} Since the Dirichlet series converges absolutely at $s_0$, we have \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s_0}| = \sum_{n=1}^{\infty} |a_n|n^{-\sigma_0} < \infty. \end{align*} Define the non-negative numerical sequence $(b_n)_{n \in \mathbb{N}}$ by \begin{align*} b_n := |a_n|n^{-\sigma_0}. \end{align*} Then $\sum_{n=1}^{\infty} b_n < \infty$. [guided] We first extract from the hypothesis a numerical series that will later dominate all tails uniformly. For each $n \in \mathbb{N}$, the complex power appearing in the Dirichlet series is defined by \begin{align*} n^{-s_0} := e^{-s_0 \log n}, \end{align*} where $\log n$ is the real logarithm of the positive integer $n$. Taking absolute values and using $|e^z| = e^{\operatorname{Re}(z)}$ for $z \in \mathbb{C}$ gives \begin{align*} |n^{-s_0}| = |e^{-s_0 \log n}| = e^{-\operatorname{Re}(s_0)\log n} = n^{-\sigma_0}, \end{align*} because $\operatorname{Re}(s_0)=\sigma_0$ by hypothesis. The assumption that the Dirichlet series converges absolutely at $s_0$ means precisely that the numerical series of absolute values converges: \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s_0}| < \infty. \end{align*} Substituting the computation of $|n^{-s_0}|$ into this series yields \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s_0}| = \sum_{n=1}^{\infty} |a_n|n^{-\sigma_0} < \infty. \end{align*} Define the non-negative numerical sequence $(b_n)_{n \in \mathbb{N}}$ by \begin{align*} b_n := |a_n|n^{-\sigma_0}. \end{align*} Then $b_n \geq 0$ for every $n \in \mathbb{N}$, and the computation above proves \begin{align*} \sum_{n=1}^{\infty} b_n < \infty. \end{align*} This sequence is the majorant that will be used on the whole shifted half-plane. [/guided] [/step] [step:Dominate every term uniformly on the shifted half-plane] Fix $\varepsilon > 0$, and define \begin{align*} H_\varepsilon := \{s \in \mathbb{C} : \operatorname{Re}(s) \geq \sigma_0 + \varepsilon\}. \end{align*} For every $s \in H_\varepsilon$ and every $n \in \mathbb{N}$, \begin{align*} |a_n n^{-s}| &= |a_n|\,|e^{-s\log n}| \\ &= |a_n|e^{-\operatorname{Re}(s)\log n} \\ &= |a_n|n^{-\operatorname{Re}(s)} \\ &\leq |a_n|n^{-(\sigma_0+\varepsilon)} \\ &= |a_n|n^{-\sigma_0}n^{-\varepsilon} \\ &\leq |a_n|n^{-\sigma_0} \\ &= b_n. \end{align*} The last inequality uses $n^{-\varepsilon} \leq 1$ for every $n \in \mathbb{N}$. [guided] Fix $\varepsilon > 0$. We want a bound on $|a_n n^{-s}|$ that is independent of $s$ as long as $s$ stays in the half-plane \begin{align*} H_\varepsilon := \{s \in \mathbb{C} : \operatorname{Re}(s) \geq \sigma_0 + \varepsilon\}. \end{align*} For $s \in H_\varepsilon$, the real part of $s$ is at least $\sigma_0+\varepsilon$. Since $n \geq 1$, increasing the exponent in $n^{-\operatorname{Re}(s)}$ decreases the value. Thus, for every $n \in \mathbb{N}$, \begin{align*} |a_n n^{-s}| &= |a_n|\,|e^{-s\log n}| \\ &= |a_n|e^{-\operatorname{Re}(s)\log n} \\ &= |a_n|n^{-\operatorname{Re}(s)} \\ &\leq |a_n|n^{-(\sigma_0+\varepsilon)} \\ &= |a_n|n^{-\sigma_0}n^{-\varepsilon}. \end{align*} Because $\varepsilon > 0$ and $n \geq 1$, we also have $n^{-\varepsilon} \leq 1$. Therefore \begin{align*} |a_n n^{-s}| \leq |a_n|n^{-\sigma_0} = b_n. \end{align*} This is the key point: the same summable sequence $(b_n)$ controls all terms simultaneously for every $s \in H_\varepsilon$. [/guided] [/step] [step:Use uniform tail estimates to prove uniform absolute convergence] For integers $M \geq N \geq 1$, define the absolute tail function \begin{align*} T_{N,M}: H_\varepsilon &\to [0,\infty) \\ s &\mapsto \sum_{n=N}^{M} |a_n n^{-s}|. \end{align*} By the uniform domination from the previous step, for every $s \in H_\varepsilon$, \begin{align*} 0 \leq T_{N,M}(s) \leq \sum_{n=N}^{M} b_n. \end{align*} Since $\sum_{n=1}^{\infty} b_n$ converges, its tails tend to $0$: for every $\delta > 0$, there exists $N_\delta \in \mathbb{N}$ such that whenever $M \geq N \geq N_\delta$, \begin{align*} \sum_{n=N}^{M} b_n < \delta. \end{align*} Hence for all $M \geq N \geq N_\delta$, \begin{align*} \sup_{s \in H_\varepsilon} \sum_{n=N}^{M} |a_n n^{-s}| \leq \sum_{n=N}^{M} b_n < \delta. \end{align*} Thus the series $\sum_{n=1}^{\infty} |a_n n^{-s}|$ satisfies the [Uniform Cauchy Criterion for Series of Functions](/theorems/???) on $H_\varepsilon$, so it converges uniformly on $H_\varepsilon$. Therefore $\sum_{n=1}^{\infty} a_n n^{-s}$ converges absolutely and uniformly on $H_\varepsilon$. [guided] We now turn the pointwise domination into uniform convergence. For integers $M \geq N \geq 1$, define the finite tail function \begin{align*} T_{N,M}: H_\varepsilon &\to [0,\infty) \\ s &\mapsto \sum_{n=N}^{M} |a_n n^{-s}|. \end{align*} The previous step gives the termwise bound $|a_n n^{-s}| \leq b_n$ for every $s \in H_\varepsilon$ and every $n \in \mathbb{N}$. Summing this bound from $n=N$ to $n=M$ gives \begin{align*} 0 \leq T_{N,M}(s) = \sum_{n=N}^{M} |a_n n^{-s}| \leq \sum_{n=N}^{M} b_n. \end{align*} The important feature is that the right-hand side no longer depends on $s$. Since $\sum_{n=1}^{\infty} b_n$ is a convergent numerical series, its tails are Cauchy. Therefore, for every $\delta > 0$, there exists $N_\delta \in \mathbb{N}$ such that for all integers $M \geq N \geq N_\delta$, \begin{align*} \sum_{n=N}^{M} b_n < \delta. \end{align*} Combining this with the tail estimate gives \begin{align*} \sup_{s \in H_\varepsilon} \sum_{n=N}^{M} |a_n n^{-s}| \leq \sum_{n=N}^{M} b_n < \delta. \end{align*} This is precisely the hypothesis of the [Uniform Cauchy Criterion for Series of Functions](/theorems/???) for the series of non-negative functions \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s}| \end{align*} on $H_\varepsilon$. Hence that series converges uniformly on $H_\varepsilon$. Since convergence of the series of absolute values at each $s$ is absolute convergence of the original Dirichlet series at that $s$, the Dirichlet series \begin{align*} \sum_{n=1}^{\infty} a_n n^{-s} \end{align*} converges absolutely and uniformly on $H_\varepsilon$. [/guided] [/step] [step:Conclude for every shifted right half-plane] The argument above was carried out for an arbitrary fixed $\varepsilon > 0$. Therefore, for every $\varepsilon > 0$, the Dirichlet series \begin{align*} \sum_{n=1}^{\infty} a_n n^{-s} \end{align*} converges absolutely and uniformly on \begin{align*} \{s \in \mathbb{C} : \operatorname{Re}(s) \geq \sigma_0 + \varepsilon\}. \end{align*} This is the desired conclusion. [/step]