[proofplan] The proof is a Dirichlet-test argument made uniform on compact subsets of the half-plane $\operatorname{Re}(s)>0$. Since $\chi$ is nonprincipal, its values over one full residue period sum to zero, so the partial sums $\sum_{n \le N}\chi(n)$ are bounded uniformly in $N$. Abel summation then rewrites the tail of $\sum \chi(n)n^{-s}$ in terms of these bounded partial sums and the derivative of $x^{-s}$, giving locally [uniform convergence](/page/Uniform%20Convergence) for $\operatorname{Re}(s)>0$. The same argument applied to the differentiated series gives local uniform convergence of the derivative series, so the locally uniform sum is holomorphic. [/proofplan]

[step:Bound the partial sums of the nonprincipal character]Let \begin{align*} A: \mathbb{N}_0 &\to \mathbb{C} \\ N &\mapsto \sum_{n=1}^{N}\chi(n), \end{align*} with the convention $A(0)=0$. We first show that $A$ is bounded. Since $\chi$ is a Dirichlet character modulo $q$, it is periodic with period $q$. Since $\chi$ is nonprincipal, its sum over a complete residue system modulo $q$ is zero: \begin{align*} \sum_{r=1}^{q}\chi(r)=0. \end{align*} Indeed, choose an integer $a$ with $\gcd(a,q)=1$ and $\chi(a)\neq 1$. Multiplication by $a$ permutes the residue classes modulo $q$, hence \begin{align*} \sum_{r=1}^{q}\chi(ar) = \sum_{r=1}^{q}\chi(r). \end{align*} By multiplicativity of $\chi$ modulo $q$, \begin{align*} \sum_{r=1}^{q}\chi(ar) = \chi(a)\sum_{r=1}^{q}\chi(r). \end{align*} Therefore \begin{align*} (1-\chi(a))\sum_{r=1}^{q}\chi(r)=0. \end{align*} Since $\chi(a)\neq 1$, the complete-period sum is zero. Now write each $N\in\mathbb{N}$ uniquely as $N=mq+r$, where $m\in\mathbb{N}_0$ and $r\in\{0,1,\dots,q-1\}$. Periodicity and the complete-period cancellation give \begin{align*} A(N) = m\sum_{j=1}^{q}\chi(j)+\sum_{j=1}^{r}\chi(j) = \sum_{j=1}^{r}\chi(j). \end{align*} Thus \begin{align*} |A(N)| \leq \sum_{j=1}^{q}|\chi(j)| \leq q \end{align*} for every $N\in\mathbb{N}_0$. Define \begin{align*} M := q. \end{align*} Then $|A(N)|\leq M$ for all $N\in\mathbb{N}_0$.[/step]

[guided]The first input is cancellation over one period. Define the partial-sum function \begin{align*} A: \mathbb{N}_0 &\to \mathbb{C} \\ N &\mapsto \sum_{n=1}^{N}\chi(n), \end{align*} with $A(0)=0$. We want a bound for $A(N)$ which is independent of $N$. Because $\chi$ is a Dirichlet character modulo $q$, it is periodic with period $q$. The essential point is that a nonprincipal character has zero average over one complete residue system. To prove this, choose an integer $a$ satisfying $\gcd(a,q)=1$ and $\chi(a)\neq 1$, which exists because $\chi$ is nonprincipal. Multiplication by $a$ permutes the residue classes modulo $q$, so \begin{align*} \sum_{r=1}^{q}\chi(ar) = \sum_{r=1}^{q}\chi(r). \end{align*} On the other hand, multiplicativity gives \begin{align*} \sum_{r=1}^{q}\chi(ar) = \chi(a)\sum_{r=1}^{q}\chi(r). \end{align*} Combining these identities yields \begin{align*} (1-\chi(a))\sum_{r=1}^{q}\chi(r)=0. \end{align*} Since $\chi(a)\neq 1$, we obtain \begin{align*} \sum_{r=1}^{q}\chi(r)=0. \end{align*} Now decompose $N=mq+r$ with $m\in\mathbb{N}_0$ and $r\in\{0,1,\dots,q-1\}$. The first $mq$ terms consist of $m$ complete periods, each summing to zero. Hence \begin{align*} A(N) = m\sum_{j=1}^{q}\chi(j)+\sum_{j=1}^{r}\chi(j) = \sum_{j=1}^{r}\chi(j). \end{align*} Therefore \begin{align*} |A(N)|\leq \sum_{j=1}^{q}|\chi(j)|\leq q. \end{align*} We set $M:=q$. This gives the uniform bound $|A(N)|\leq M$ for all $N\in\mathbb{N}_0$.[/guided]

[step:Use Abel summation to obtain locally uniform convergence on $\operatorname{Re}(s)>0$]Let $K\subset H$ be compact. Since $K$ is compact and contained in $H$, define \begin{align*} \sigma_K := \inf_{s\in K}\operatorname{Re}(s), \end{align*} so that $\sigma_K>0$. For integers $1\leq m\leq N$, define \begin{align*} f_s: [1,\infty) &\to \mathbb{C} \\ x &\mapsto x^{-s}=\exp(-s\log x). \end{align*} The function $f_s$ is continuously differentiable, and \begin{align*} f_s'(x)=-s x^{-s-1}. \end{align*} Abel summation gives \begin{align*} \sum_{n=m}^{N}\chi(n)n^{-s} = A(N)N^{-s}-A(m-1)m^{-s} + \int_m^N A(\lfloor x\rfloor)sx^{-s-1}\,d\mathcal{L}^1(x). \end{align*} Taking absolute values and using $|A(k)|\leq M$ gives, for every $s\in K$, \begin{align*} \left|\sum_{n=m}^{N}\chi(n)n^{-s}\right| &\leq M N^{-\operatorname{Re}(s)}+M m^{-\operatorname{Re}(s)} +M|s|\int_m^N x^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(x)\\ &\leq M m^{-\sigma_K}+M m^{-\sigma_K} +M\sup_{s\in K}|s|\int_m^\infty x^{-\sigma_K-1}\,d\mathcal{L}^1(x)\\ &= 2M m^{-\sigma_K} + \frac{M\sup_{s\in K}|s|}{\sigma_K}m^{-\sigma_K}. \end{align*} The right-hand side tends to $0$ as $m\to\infty$, uniformly in $N\geq m$ and $s\in K$. Hence the partial sums of \begin{align*} \sum_{n=1}^{\infty}\chi(n)n^{-s} \end{align*} form a uniformly [Cauchy sequence](/page/Cauchy%20Sequence) on $K$. Since $K\subset H$ was arbitrary, the series converges locally uniformly on $H$.[/step]

[guided]Fix a compact set $K\subset H$. The goal is to prove uniform convergence on $K$, because local uniform convergence means exactly uniform convergence on every compact subset. Since $K$ lies inside the open half-plane $H$, its real parts are bounded away from zero. Define \begin{align*} \sigma_K := \inf_{s\in K}\operatorname{Re}(s). \end{align*} Compactness of $K$ and the inclusion $K\subset H$ imply $\sigma_K>0$. For each $s\in K$, define the function \begin{align*} f_s: [1,\infty) &\to \mathbb{C} \\ x &\mapsto x^{-s}=\exp(-s\log x). \end{align*} This is continuously differentiable, and its derivative is \begin{align*} f_s'(x)=-s x^{-s-1}. \end{align*} We apply Abel summation to the finite sum from $m$ to $N$. Since $A(k)=\sum_{n=1}^{k}\chi(n)$, Abel summation gives \begin{align*} \sum_{n=m}^{N}\chi(n)n^{-s} = A(N)N^{-s}-A(m-1)m^{-s} + \int_m^N A(\lfloor x\rfloor)sx^{-s-1}\,d\mathcal{L}^1(x). \end{align*} The measure is the one-dimensional Lebesgue measure $\mathcal{L}^1$ on $[m,N]$. Now use the boundedness of the partial sums. From the previous step, $|A(k)|\leq M$ for every $k\in\mathbb{N}_0$. Therefore, for $s\in K$, \begin{align*} \left|\sum_{n=m}^{N}\chi(n)n^{-s}\right| &\leq M N^{-\operatorname{Re}(s)}+M m^{-\operatorname{Re}(s)} +M|s|\int_m^N x^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(x). \end{align*} Since $\operatorname{Re}(s)\geq \sigma_K$ and $N\geq m$, we have \begin{align*} N^{-\operatorname{Re}(s)}\leq m^{-\sigma_K}, \qquad m^{-\operatorname{Re}(s)}\leq m^{-\sigma_K}. \end{align*} Also, \begin{align*} \int_m^N x^{-\operatorname{Re}(s)-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_{n=m}^{N}\chi(n)n^{-s}\right| \leq 2M m^{-\sigma_K} + \frac{M\sup_{s\in K}|s|}{\sigma_K}m^{-\sigma_K}. \end{align*} The right-hand side tends to $0$ as $m\to\infty$, and it does not depend on $N$ or on the particular $s\in K$. Hence the tails are uniformly small on $K$, so the series is uniformly Cauchy on $K$. Therefore \begin{align*} \sum_{n=1}^{\infty}\chi(n)n^{-s} \end{align*} converges uniformly on $K$. Since $K\subset H$ was arbitrary, the convergence is locally uniform on $H$.[/guided]

[step:Prove holomorphy by locally uniform convergence of the differentiated series]For each $n\in\mathbb{N}$, define \begin{align*} g_n: H &\to \mathbb{C} \\ s &\mapsto \chi(n)n^{-s}. \end{align*} Each $g_n$ is entire as a function of $s$, and \begin{align*} g_n'(s)=-\chi(n)(\log n)n^{-s}. \end{align*} We show that the derivative series converges locally uniformly on $H$. Fix again a compact set $K\subset H$ and let $\sigma_K>0$ be as above. For $s\in K$, apply Abel summation to the coefficients $\chi(n)$ and the function \begin{align*} h_s: [1,\infty) &\to \mathbb{C} \\ x &\mapsto (\log x)x^{-s}. \end{align*} For $x>1$, \begin{align*} h_s'(x) = x^{-s-1}(1-s\log x). \end{align*} Thus, for $2\leq m\leq N$, \begin{align*} \left|\sum_{n=m}^{N}\chi(n)(\log n)n^{-s}\right| &\leq M(\log N)N^{-\operatorname{Re}(s)} + M(\log m)m^{-\operatorname{Re}(s)}\\ &\quad+ M\int_m^N x^{-\operatorname{Re}(s)-1}|1-s\log x|\,d\mathcal{L}^1(x)\\ &\leq 2M(\log m)m^{-\sigma_K} + M\int_m^\infty x^{-\sigma_K-1}\left(1+\sup_{s\in K}|s|\log x\right)\,d\mathcal{L}^1(x). \end{align*} The final expression tends to $0$ as $m\to\infty$, because $\sigma_K>0$. Hence \begin{align*} \sum_{n=1}^{\infty}-\chi(n)(\log n)n^{-s} \end{align*} converges locally uniformly on $H$. Let \begin{align*} S_N: H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{N}\chi(n)n^{-s}. \end{align*} The functions $S_N$ are holomorphic, the sequence $(S_N)$ converges locally uniformly on $H$ by the previous step, and the derivative sequence $(S_N')$ converges locally uniformly on $H$ by the preceding estimate. Therefore the locally uniform limit \begin{align*} L(\cdot,\chi): H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\chi(n)n^{-s} \end{align*} is holomorphic on $H$, with derivative \begin{align*} L'(s,\chi)=\sum_{n=1}^{\infty}-\chi(n)(\log n)n^{-s}. \end{align*}[/step]

[guided]We have proved local uniform convergence of the original Dirichlet series. To prove holomorphy directly, we also prove local uniform convergence of the termwise derivative series. For each $n\in\mathbb{N}$, define \begin{align*} g_n: H &\to \mathbb{C} \\ s &\mapsto \chi(n)n^{-s}. \end{align*} Since $n^{-s}=\exp(-s\log n)$ and $\log n$ is a real constant, $g_n$ is entire as a function of $s$. Its derivative is \begin{align*} g_n'(s)=-\chi(n)(\log n)n^{-s}. \end{align*} Fix a compact set $K\subset H$, and define \begin{align*} \sigma_K := \inf_{s\in K}\operatorname{Re}(s)>0. \end{align*} We estimate the tails of the derivative series. Apply Abel summation again, now to the function \begin{align*} h_s: [1,\infty) &\to \mathbb{C} \\ x &\mapsto (\log x)x^{-s}. \end{align*} For $x>1$, differentiation gives \begin{align*} h_s'(x)=x^{-s-1}(1-s\log x). \end{align*} Using $|A(k)|\leq M$, Abel summation gives, for $2\leq m\leq N$, \begin{align*} \left|\sum_{n=m}^{N}\chi(n)(\log n)n^{-s}\right| &\leq M(\log N)N^{-\operatorname{Re}(s)} + M(\log m)m^{-\operatorname{Re}(s)}\\ &\quad+ M\int_m^N x^{-\operatorname{Re}(s)-1}|1-s\log x|\,d\mathcal{L}^1(x). \end{align*} Since $\operatorname{Re}(s)\geq \sigma_K$, and since the function $x\mapsto (\log x)x^{-\sigma_K}$ tends to $0$ as $x\to\infty$, the two boundary terms are bounded by a quantity tending to $0$ uniformly in $s\in K$. More explicitly, for all sufficiently large $m$ and all $N\geq m$, \begin{align*} (\log N)N^{-\operatorname{Re}(s)} \leq (\log m)m^{-\sigma_K} \end{align*} after increasing the initial threshold if necessary; this is enough for tail convergence. For the integral term, \begin{align*} \int_m^N x^{-\operatorname{Re}(s)-1}|1-s\log x|\,d\mathcal{L}^1(x) \leq \int_m^\infty x^{-\sigma_K-1}\left(1+\sup_{s\in K}|s|\log x\right)\,d\mathcal{L}^1(x). \end{align*} The improper integral on the right tends to $0$ as $m\to\infty$, because $\sigma_K>0$ and both \begin{align*} \int_m^\infty x^{-\sigma_K-1}\,d\mathcal{L}^1(x) \quad\text{and}\quad \int_m^\infty x^{-\sigma_K-1}\log x\,d\mathcal{L}^1(x) \end{align*} converge and have tails tending to zero. Hence the derivative series \begin{align*} \sum_{n=1}^{\infty}-\chi(n)(\log n)n^{-s} \end{align*} converges locally uniformly on $H$. Now define the finite partial sums \begin{align*} S_N: H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{N}\chi(n)n^{-s}. \end{align*} Each $S_N$ is holomorphic because it is a finite sum of entire functions. We have proved that $S_N$ converges locally uniformly to \begin{align*} L(\cdot,\chi): H &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\chi(n)n^{-s}, \end{align*} and that $S_N'$ converges locally uniformly to \begin{align*} s\mapsto \sum_{n=1}^{\infty}-\chi(n)(\log n)n^{-s}. \end{align*} The standard local theorem on termwise differentiation of holomorphic functions then gives that $L(\cdot,\chi)$ is holomorphic on $H$ and that \begin{align*} L'(s,\chi)=\sum_{n=1}^{\infty}-\chi(n)(\log n)n^{-s}. \end{align*}[/guided]

[step:Conclude holomorphy at $s=1$] Since \begin{align*} 1\in H=\{s\in\mathbb{C}:\operatorname{Re}(s)>0\}, \end{align*} the holomorphy of $L(\cdot,\chi)$ on $H$ implies that $L(s,\chi)$ is holomorphic in a neighbourhood of $s=1$. This proves the asserted holomorphy at one. The argument proves holomorphy only; it does not imply $L(1,\chi)\neq 0$. [/step]