[proofplan] We first prove that the partial sums of a nonprincipal Dirichlet character are uniformly bounded, using periodicity and the vanishing of the sum over one complete residue system. We then derive Abel summation for the tail $\sum_{M<n\leq N}\chi(n)n^{-s}$ with all boundary terms explicit. The boundedness of the partial sums makes the Abel summation formula uniformly Cauchy on compact subsets of the half-plane $\operatorname{Re}(s)>0$. Finally, the locally uniform limit of the holomorphic partial sums is holomorphic. [/proofplan] [step:Define the common half-plane and measure notation] Throughout the proof, define the open right half-plane $H \subset \mathbb{C}$ by \begin{align*} H := \{s\in\mathbb{C}: \operatorname{Re}(s)>0\}, \end{align*} and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. [/step] [step:Bound the character sums over initial intervals] Define the complete-period character sum $S_\chi \in \mathbb{C}$ by \begin{align*} S_\chi := \sum_{a=1}^{q} \chi(a). \end{align*} Because $\chi$ is nonprincipal, there exists an integer $b \in \mathbb{Z}$ with $\gcd(b,q)=1$ and $\chi(b) \neq 1$. Multiplication by $b$ permutes the residue classes modulo $q$, so \begin{align*} S_\chi = \sum_{a=1}^{q} \chi(a) = \sum_{a=1}^{q} \chi(ba) = \chi(b)\sum_{a=1}^{q}\chi(a) = \chi(b)S_\chi. \end{align*} Since $\chi(b)\neq 1$, this gives $S_\chi=0$. Define the partial-sum function \begin{align*} A: [1,\infty) &\to \mathbb{C} \\ x &\mapsto \sum_{1\leq n\leq x}\chi(n), \end{align*} where the condition $1\leq n\leq x$ means $1\leq n\leq \lfloor x\rfloor$. If $\lfloor x\rfloor = mq+r$ with $m\in \mathbb{N}\cup\{0\}$ and $r\in\{0,1,\dots,q-1\}$, then the contribution of the $m$ complete blocks of length $q$ is $mS_\chi=0$, and therefore \begin{align*} A(x)=\sum_{a=1}^{r}\chi(a). \end{align*} Since $|\chi(a)|\leq 1$ for every $a\in\mathbb{Z}$, we obtain \begin{align*} |A(x)|\leq q \end{align*} for every $x\geq 1$. [guided] The key input is cancellation over one full period. Let \begin{align*} S_\chi := \sum_{a=1}^{q}\chi(a) \end{align*} be the sum of $\chi$ over one complete set of residues modulo $q$. Since $\chi$ is nonprincipal, it is not identically $1$ on the reduced residue classes modulo $q$. Hence there is an integer $b$ with $\gcd(b,q)=1$ and $\chi(b)\neq 1$. Multiplication by such a $b$ is a bijection on the residue classes modulo $q$. Thus reindexing the complete residue sum by $a\mapsto ba$ gives \begin{align*} S_\chi = \sum_{a=1}^{q}\chi(a) = \sum_{a=1}^{q}\chi(ba). \end{align*} Because $\chi$ is multiplicative as a Dirichlet character, \begin{align*} \sum_{a=1}^{q}\chi(ba) = \chi(b)\sum_{a=1}^{q}\chi(a) = \chi(b)S_\chi. \end{align*} Therefore $S_\chi=\chi(b)S_\chi$. Since $\chi(b)\neq 1$, subtracting gives $(1-\chi(b))S_\chi=0$, so $S_\chi=0$. Now define \begin{align*} A: [1,\infty) &\to \mathbb{C} \\ x &\mapsto \sum_{1\leq n\leq x}\chi(n), \end{align*} where the finite sum is over integers $n$ with $1\leq n\leq \lfloor x\rfloor$. Write $\lfloor x\rfloor=mq+r$ with $m\in\mathbb{N}\cup\{0\}$ and $r\in\{0,1,\dots,q-1\}$. The first $mq$ terms split into $m$ complete periods, each of which has sum $S_\chi=0$. Only the remaining $r$ terms can contribute: \begin{align*} A(x)=\sum_{a=1}^{r}\chi(a). \end{align*} Since every value of a Dirichlet character has absolute value at most $1$, this gives \begin{align*} |A(x)|\leq \sum_{a=1}^{r}|\chi(a)|\leq r\leq q \end{align*} for every $x\geq 1$. Thus the partial sums of $\chi$ are bounded independently of $x$. [/guided] [/step] [step:Express finite tails by Abel summation] Let $M,N\in\mathbb{N}$ with $M<N$, and let $s\in H$. Define the function \begin{align*} f_s: [M,N] &\to \mathbb{C} \\ x &\mapsto x^{-s}:=\exp(-s\log x), \end{align*} where $\log x$ is the real logarithm. Since $f_s$ is continuously differentiable on $[M,N]$ and \begin{align*} f_s'(x)=-s x^{-s-1}, \end{align*} we compute \begin{align*} s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x) &= \sum_{n=M}^{N-1} s A(n)\int_n^{n+1}x^{-s-1}\,d\mathcal{L}^1(x) \\ &= \sum_{n=M}^{N-1} A(n)\bigl(n^{-s}-(n+1)^{-s}\bigr). \end{align*} Therefore \begin{align*} &A(N)N^{-s}-A(M)M^{-s} +s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x) \\ &\qquad = A(N)N^{-s}-A(M)M^{-s} +\sum_{n=M}^{N-1}A(n)\bigl(n^{-s}-(n+1)^{-s}\bigr) \\ &\qquad = \sum_{n=M+1}^{N}\bigl(A(n)-A(n-1)\bigr)n^{-s}. \end{align*} Since $A(n)-A(n-1)=\chi(n)$ for every integer $n\geq 2$, we obtain the Abel summation identity \begin{align*} \sum_{M<n\leq N}\frac{\chi(n)}{n^s} = A(N)N^{-s}-A(M)M^{-s} +s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} [guided] We want to replace the oscillating tail of $\chi(n)n^{-s}$ by a formula involving the bounded partial sums $A(x)$. Fix integers $M<N$ and a complex number $s\in H$. Define \begin{align*} f_s: [M,N] &\to \mathbb{C} \\ x &\mapsto x^{-s}:=\exp(-s\log x), \end{align*} where $\log x$ is the real logarithm. This function is continuously differentiable on $[M,N]$, and differentiation gives \begin{align*} f_s'(x)=-s x^{-s-1}. \end{align*} The function $A$ is constant on each half-open interval $[n,n+1)$ for $n=M,\dots,N-1$, up to endpoints of $\mathcal{L}^1$-measure zero. Hence \begin{align*} s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x) &= \sum_{n=M}^{N-1}sA(n)\int_n^{n+1}x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} Using $f_s'(x)=-s x^{-s-1}$, each integral becomes \begin{align*} s\int_n^{n+1}x^{-s-1}\,d\mathcal{L}^1(x) = n^{-s}-(n+1)^{-s}. \end{align*} Therefore \begin{align*} s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x) = \sum_{n=M}^{N-1}A(n)\bigl(n^{-s}-(n+1)^{-s}\bigr). \end{align*} Now add the boundary terms. The expression \begin{align*} A(N)N^{-s}-A(M)M^{-s} +\sum_{n=M}^{N-1}A(n)\bigl(n^{-s}-(n+1)^{-s}\bigr) \end{align*} telescopes: the coefficient of $n^{-s}$ for $M<n<N$ is $A(n)-A(n-1)$, the coefficient of $N^{-s}$ is also $A(N)-A(N-1)$, and the $M^{-s}$ terms cancel. Hence the whole expression equals \begin{align*} \sum_{n=M+1}^{N}\bigl(A(n)-A(n-1)\bigr)n^{-s}. \end{align*} By definition of $A$, we have $A(n)-A(n-1)=\chi(n)$ for every integer $n\geq 2$. Thus \begin{align*} \sum_{M<n\leq N}\frac{\chi(n)}{n^s} = A(N)N^{-s}-A(M)M^{-s} +s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} This is Abel summation in the precise form needed for the convergence estimate. [/guided] [/step] [step:Prove compact-uniform convergence in the half-plane] Let $K\subset H$ be compact. Define \begin{align*} \sigma_K := \min_{s\in K}\operatorname{Re}(s), \qquad R_K := \max_{s\in K}|s|. \end{align*} Since $K\subset H$ is compact, $\sigma_K>0$ and $R_K<\infty$. For $s\in K$ and $M<N$, the Abel summation identity and the bound $|A(x)|\leq q$ give \begin{align*} \left|\sum_{M<n\leq N}\frac{\chi(n)}{n^s}\right| &\leq qN^{-\operatorname{Re}(s)}+qM^{-\operatorname{Re}(s)} +|s|q\int_M^N x^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(x) \\ &\leq qN^{-\sigma_K}+qM^{-\sigma_K} +R_Kq\int_M^N x^{-\sigma_K-1}\,d\mathcal{L}^1(x) \\ &\leq 2qM^{-\sigma_K} +\frac{R_Kq}{\sigma_K}M^{-\sigma_K}. \end{align*} The final expression tends to $0$ as $M\to\infty$, independently of $N$ and $s\in K$. Hence the partial sums \begin{align*} S_N: K &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{N}\frac{\chi(n)}{n^s} \end{align*} form a uniformly [Cauchy sequence](/page/Cauchy%20Sequence) on $K$. Therefore the Dirichlet series converges uniformly on $K$. Since $K\subset H$ was arbitrary, the convergence is locally uniform on $H$. [guided] Fix a compact set $K\subset H$. To prove locally [uniform convergence](/page/Uniform%20Convergence), it is enough to prove that the tails are uniformly small for all $s\in K$. Define \begin{align*} \sigma_K := \min_{s\in K}\operatorname{Re}(s), \qquad R_K := \max_{s\in K}|s|. \end{align*} The compactness of $K$ and the inclusion $K\subset H$ imply $\sigma_K>0$. Also $R_K<\infty$. Take $s\in K$ and integers $M<N$. Abel summation gives \begin{align*} \sum_{M<n\leq N}\frac{\chi(n)}{n^s} = A(N)N^{-s}-A(M)M^{-s} +s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} We now estimate each term. Since $|A(x)|\leq q$ and $|x^{-s}|=x^{-\operatorname{Re}(s)}$ for $x>0$, \begin{align*} |A(N)N^{-s}|\leq qN^{-\operatorname{Re}(s)}, \qquad |A(M)M^{-s}|\leq qM^{-\operatorname{Re}(s)}. \end{align*} For the integral term, the triangle inequality for Lebesgue integrals gives \begin{align*} \left|s\int_M^N A(x)x^{-s-1}\,d\mathcal{L}^1(x)\right| \leq |s|q\int_M^N x^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(x). \end{align*} Since $\operatorname{Re}(s)\geq \sigma_K$ and $|s|\leq R_K$ for every $s\in K$, we get \begin{align*} \left|\sum_{M<n\leq N}\frac{\chi(n)}{n^s}\right| &\leq qN^{-\sigma_K}+qM^{-\sigma_K} +R_Kq\int_M^N x^{-\sigma_K-1}\,d\mathcal{L}^1(x). \end{align*} Because $N>M$, the first boundary term satisfies $N^{-\sigma_K}\leq M^{-\sigma_K}$. Also \begin{align*} \int_M^N x^{-\sigma_K-1}\,d\mathcal{L}^1(x) \leq \int_M^\infty x^{-\sigma_K-1}\,d\mathcal{L}^1(x) = \frac{1}{\sigma_K}M^{-\sigma_K}. \end{align*} Thus \begin{align*} \left|\sum_{M<n\leq N}\frac{\chi(n)}{n^s}\right| \leq \left(2q+\frac{R_Kq}{\sigma_K}\right)M^{-\sigma_K}. \end{align*} The right-hand side tends to $0$ as $M\to\infty$ and does not depend on $N$ or on $s\in K$. Hence the partial sums are uniformly Cauchy on $K$, so the series converges uniformly on $K$. Since every compact subset of $H$ can be used as $K$, the convergence is locally uniform on $H$. [/guided] [/step] [step:Pass holomorphy to the locally uniform limit] For each $N\in\mathbb{N}$, define \begin{align*} S_N: H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{N}\chi(n)\exp(-s\log n). \end{align*} Each summand $s\mapsto \chi(n)\exp(-s\log n)$ is entire, so each $S_N$ is holomorphic on $H$. The preceding step shows that $(S_N)$ converges locally uniformly on $H$ to \begin{align*} L(\cdot,\chi): H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}. \end{align*} By the Weierstrass theorem on locally uniform limits of holomorphic functions, applied on the [open set](/page/Open%20Set) $H$ to the holomorphic functions $(S_N)$ and using the locally uniform convergence established above, the limit function $L(\cdot,\chi)$ is holomorphic on $H$. This proves both the asserted compact-uniform convergence and the holomorphy of the Dirichlet $L$-series in the half-plane $\operatorname{Re}(s)>0$. [/step]