[proofplan] We first recover the Euler product for $L(s,\chi)$ in the half-plane $\operatorname{Re}(s)>1$ from absolute convergence and complete multiplicativity of $\chi$. We then take logarithms by expanding each local factor through the absolutely and locally uniformly convergent [power series](/page/Power%20Series) for $-\log(1-z)$. Differentiating the resulting locally uniformly convergent series gives a sum indexed by prime powers, and reindexing those prime powers by the von Mangoldt function gives the stated Dirichlet series. [/proofplan] [step:Recover the Euler product from absolute convergence and complete multiplicativity] Let \begin{align*} H := \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} \end{align*} denote the half-plane of absolute convergence, and let $\mathcal{P}$ denote the set of prime numbers. For $s \in H$, the Dirichlet series defining $L(s,\chi)$ converges absolutely because $|\chi(n)| \leq 1$ for every $n \in \mathbb{N}$ and \begin{align*} \sum_{n=1}^{\infty}\left|\frac{\chi(n)}{n^s}\right| \leq \sum_{n=1}^{\infty}\frac{1}{n^{\operatorname{Re}(s)}} < \infty. \end{align*} For a finite set $P \subset \mathcal{P}$, define \begin{align*} E_P(\cdot): H &\to \mathbb{C} \\ s &\mapsto \prod_{p \in P}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} Since $|\chi(p)p^{-s}| \leq p^{-\operatorname{Re}(s)}<1$, each factor is well-defined and expands as \begin{align*} \left(1-\frac{\chi(p)}{p^s}\right)^{-1} = \sum_{k=0}^{\infty}\frac{\chi(p)^k}{p^{ks}}. \end{align*} Multiplying the finitely many absolutely convergent geometric series and using complete multiplicativity of $\chi$, we obtain \begin{align*} E_P(s) = \sum_{\substack{n \in \mathbb{N} \\ \text{all prime divisors of } n \text{ lie in } P}} \frac{\chi(n)}{n^s}. \end{align*} As $P$ increases through finite subsets of $\mathcal{P}$, these partial Euler products converge to the absolutely convergent Dirichlet series. Therefore \begin{align*} L(s,\chi) = \prod_{p \in \mathcal{P}}\left(1-\frac{\chi(p)}{p^s}\right)^{-1} \end{align*} for every $s \in H$. [guided] The first point is that the half-plane $\operatorname{Re}(s)>1$ is exactly where the comparison with the ordinary $p$-series is available. Since a Dirichlet character satisfies $|\chi(n)| \leq 1$, for every $s \in H$ we have \begin{align*} \sum_{n=1}^{\infty}\left|\frac{\chi(n)}{n^s}\right| \leq \sum_{n=1}^{\infty}\frac{1}{n^{\operatorname{Re}(s)}} < \infty. \end{align*} Thus all rearrangements below are controlled by absolute convergence. Let $\mathcal{P}$ be the set of prime numbers. For a finite set $P \subset \mathcal{P}$, define \begin{align*} E_P(\cdot): H &\to \mathbb{C} \\ s &\mapsto \prod_{p \in P}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} For each $p \in P$, the inequality $|\chi(p)p^{-s}| \leq p^{-\operatorname{Re}(s)}<1$ permits the geometric expansion \begin{align*} \left(1-\frac{\chi(p)}{p^s}\right)^{-1} = \sum_{k=0}^{\infty}\frac{\chi(p)^k}{p^{ks}}. \end{align*} Multiplying these finitely many series is legitimate because each one is absolutely convergent. Complete multiplicativity of $\chi$ gives $\chi(p_1^{k_1}\cdots p_m^{k_m})=\chi(p_1)^{k_1}\cdots\chi(p_m)^{k_m}$, so the product becomes \begin{align*} E_P(s) = \sum_{\substack{n \in \mathbb{N} \\ \text{all prime divisors of } n \text{ lie in } P}} \frac{\chi(n)}{n^s}. \end{align*} Letting $P$ increase through finite subsets of the primes exhausts the positive integers by unique factorisation. Since the full Dirichlet series is absolutely convergent, these finite-prime partial sums converge to $L(s,\chi)$. Hence \begin{align*} L(s,\chi) = \prod_{p \in \mathcal{P}}\left(1-\frac{\chi(p)}{p^s}\right)^{-1} \end{align*} for every $s \in H$. [/guided] [/step] [step:Expand the logarithm as a locally uniformly convergent prime power series] Fix $\delta>0$, and define the closed half-plane \begin{align*} H_\delta := \{s \in \mathbb{C} : \operatorname{Re}(s)\geq 1+\delta\}. \end{align*} For $p \in \mathcal{P}$, $k \in \mathbb{N}$, and $s \in H_\delta$, \begin{align*} \left|\frac{\chi(p)^k}{k p^{ks}}\right| \leq \frac{1}{k p^{k(1+\delta)}}. \end{align*} Moreover, \begin{align*} \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{1}{k p^{k(1+\delta)}} \leq \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{1}{p^{k(1+\delta)}} \leq \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^{k(1+\delta)}} <\infty. \end{align*} Thus the series \begin{align*} F(\cdot): H &\to \mathbb{C} \\ s &\mapsto \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}} \end{align*} converges absolutely and locally uniformly on $H$. For each prime $p$ and each $s \in H$, the power series identity \begin{align*} -\log(1-z)=\sum_{k=1}^{\infty}\frac{z^k}{k}, \qquad |z|<1, \end{align*} applied to $z=\chi(p)p^{-s}$ gives \begin{align*} \exp\left(\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}\right) = \left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} Combining this identity over primes with the locally [uniform convergence](/page/Uniform%20Convergence) of $F$ and the Euler product from the previous step yields \begin{align*} L(s,\chi)=\exp(F(s)). \end{align*} In particular, $L(s,\chi)\neq 0$ for every $s \in H$. [guided] The purpose of this step is to replace the Euler product by a sum, because logarithmic differentiation is easier on a sum than on a product. We must justify that the double series over primes and powers is locally uniformly convergent. Fix $\delta>0$ and define \begin{align*} H_\delta := \{s \in \mathbb{C} : \operatorname{Re}(s)\geq 1+\delta\}. \end{align*} For every $s \in H_\delta$, every prime $p$, and every $k \in \mathbb{N}$, \begin{align*} \left|\frac{\chi(p)^k}{k p^{ks}}\right| \leq \frac{1}{k p^{k(1+\delta)}}. \end{align*} The majorant is summable: \begin{align*} \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{1}{k p^{k(1+\delta)}} \leq \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{1}{p^{k(1+\delta)}} \leq \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^{k(1+\delta)}} <\infty. \end{align*} Therefore the Weierstrass majorant argument gives absolute and uniform convergence on $H_\delta$. Since every compact subset of $H$ is contained in some $H_\delta$, the convergence is locally uniform on $H$. Now define \begin{align*} F(\cdot): H &\to \mathbb{C} \\ s &\mapsto \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} For each fixed prime $p$, the inequality $|\chi(p)p^{-s}|<1$ allows the power series expansion \begin{align*} -\log(1-z)=\sum_{k=1}^{\infty}\frac{z^k}{k} \end{align*} with $z=\chi(p)p^{-s}$. Hence \begin{align*} \exp\left(\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}\right) = \left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} Multiplying over all primes is justified by the absolute and locally uniform convergence just proved, and the Euler product already established gives \begin{align*} L(s,\chi)=\exp(F(s)). \end{align*} This identity also proves that $L(s,\chi)$ has no zeros in $H$, because the exponential function never vanishes. [/guided] [/step] [step:Differentiate the locally uniformly convergent logarithmic series] For fixed $\delta>0$, the derivative of the summand \begin{align*} s \mapsto \frac{\chi(p)^k}{k p^{ks}} \end{align*} is \begin{align*} s \mapsto -(\log p)\frac{\chi(p)^k}{p^{ks}}. \end{align*} For $s \in H_\delta$, \begin{align*} \left|(\log p)\frac{\chi(p)^k}{p^{ks}}\right| \leq \frac{\log p}{p^{k(1+\delta)}}. \end{align*} The majorant is summable because \begin{align*} \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty}\frac{\log p}{p^{k(1+\delta)}} \leq \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{\log n}{n^{k(1+\delta)}} <\infty. \end{align*} Thus the differentiated series converges locally uniformly on $H$, and termwise differentiation gives \begin{align*} F'(s) = -\sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty} \frac{(\log p)\chi(p)^k}{p^{ks}}. \end{align*} Since $L(s,\chi)=\exp(F(s))$, the chain rule gives \begin{align*} \frac{L'}{L}(s,\chi)=F'(s). \end{align*} Therefore \begin{align*} -\frac{L'}{L}(s,\chi) = \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty} \frac{(\log p)\chi(p)^k}{p^{ks}}. \end{align*} [/step] [step:Reindex the prime power series by the von Mangoldt function] If $n=p^k$ for a prime $p$ and $k \in \mathbb{N}$, then complete multiplicativity gives $\chi(n)=\chi(p)^k$, and the definition of $\Lambda$ gives $\Lambda(n)=\log p$. If $n$ is not a prime power, then $\Lambda(n)=0$. Hence the double sum over prime powers is exactly the Dirichlet series \begin{align*} \sum_{p \in \mathcal{P}}\sum_{k=1}^{\infty} \frac{(\log p)\chi(p)^k}{p^{ks}} = \sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s}. \end{align*} Combining this identity with the previous step gives, for every $s \in H$, \begin{align*} -\frac{L'}{L}(s,\chi)=\sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s}. \end{align*} This is the desired logarithmic derivative formula. [/step]