[proofplan] We use the logarithm specified in the statement: each Euler factor is expanded by the branch determined by the [power series](/page/Power%20Series) for $-\log(1-z)$, valid because $|\chi(p)p^{-s}|<1$ for every prime $p$ and every $s>1$. We first justify that the resulting sum of local logarithms converges absolutely for fixed $s>1$, so the logarithmic Euler product is well-defined. The terms with exponent $k=1$ give exactly the prime sum $\sum_p \chi(p)p^{-s}$, and the remaining prime-power terms with $k\geq 2$ are bounded absolutely and uniformly for $1<s\leq 2$ by a convergent numerical series dominated by $\sum_p p^{-2}$. [/proofplan] [step:Expand the logarithm of each Euler factor] Fix $s>1$. For each rational prime $p$, define \begin{align*} z_p(s) := \chi(p)p^{-s} \in \mathbb{C}. \end{align*} Since $\chi$ is a Dirichlet character, $|\chi(p)|\leq 1$, and therefore \begin{align*} |z_p(s)| \leq p^{-s} < 1. \end{align*} Thus the logarithmic power series gives \begin{align*} -\log(1-z_p(s))=\sum_{k=1}^{\infty}\frac{z_p(s)^k}{k} =\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} We now justify that the sum of local logarithms is absolutely convergent. Since $|z_p(s)|\leq p^{-s}\leq 2^{-s}<1/2$, the triangle inequality applied to the logarithmic power series gives \begin{align*} \left|-\log(1-z_p(s))\right| &\leq \sum_{k=1}^{\infty}\frac{|z_p(s)|^k}{k} \\ &\leq \sum_{k=1}^{\infty}|z_p(s)|^k \\ &=\frac{|z_p(s)|}{1-|z_p(s)|} \\ &\leq 2p^{-s}. \end{align*} The numerical series $\sum_{n=2}^{\infty}n^{-s}$ converges because $s>1$, and the set of rational primes is contained in $\{2,3,4,\dots\}$; hence $\sum_p p^{-s}$ converges. Therefore $\sum_p |\log(1-z_p(s))|$ converges, so the logarithmic Euler product is well-defined by the absolutely convergent series of local logarithms. Using the definition of $\log L(s,\chi)$ from the statement and substituting the local power-series expansion gives \begin{align*} \log L(s,\chi) =\sum_p -\log\left(1-\chi(p)p^{-s}\right) =\sum_p \sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} [guided] Fix a real number $s>1$. For each rational prime $p$, we isolate the quantity appearing in the Euler factor by defining \begin{align*} z_p(s) := \chi(p)p^{-s} \in \mathbb{C}. \end{align*} A Dirichlet character satisfies $|\chi(n)|\leq 1$ for every $n\in\mathbb{N}$, so in particular \begin{align*} |z_p(s)| = |\chi(p)|p^{-s}\leq p^{-s}<1. \end{align*} This strict inequality is exactly what is needed to use the logarithmic power series \begin{align*} -\log(1-z)=\sum_{k=1}^{\infty}\frac{z^k}{k}, \qquad |z|<1. \end{align*} Applying this with $z=z_p(s)$ gives, for each prime $p$, \begin{align*} -\log\left(1-\chi(p)p^{-s}\right) =\sum_{k=1}^{\infty}\frac{\left(\chi(p)p^{-s}\right)^k}{k} =\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} We must also check that the sum of these local logarithms converges absolutely. The bound $|z_p(s)|\leq p^{-s}\leq 2^{-s}<1/2$ gives \begin{align*} \left|-\log(1-z_p(s))\right| &\leq \sum_{k=1}^{\infty}\frac{|z_p(s)|^k}{k} \\ &\leq \sum_{k=1}^{\infty}|z_p(s)|^k \\ &=\frac{|z_p(s)|}{1-|z_p(s)|} \\ &\leq 2p^{-s}. \end{align*} The comparison series $\sum_p p^{-s}$ converges: the primes form a subset of $\{2,3,4,\dots\}$, and $\sum_{n=2}^{\infty}n^{-s}$ converges for $s>1$. Hence $\sum_p |\log(1-z_p(s))|$ converges. This verifies that the logarithmic Euler product in the statement is well-defined as the absolutely convergent sum of the chosen local logarithms, and therefore \begin{align*} \log L(s,\chi) =\sum_p -\log\left(1-\chi(p)p^{-s}\right) =\sum_p \sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} The point of this expansion is that the term $k=1$ is the desired prime sum, while all terms $k\geq 2$ will form a uniformly bounded error. [/guided] [/step] [step:Separate the prime term from the higher prime-power terms] Define the remainder function $R:(1,\infty)\to\mathbb{C}$ by \begin{align*} R(s):=\sum_p\sum_{k=2}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} Separating the $k=1$ term in the preceding expansion gives \begin{align*} \log L(s,\chi) =\sum_p \frac{\chi(p)}{p^s}+R(s) =\sum_p \chi(p)p^{-s}+R(s). \end{align*} Thus it remains to prove that $R(s)$ remains bounded as $s\to 1^+$. [guided] We now isolate the part of the expansion that is not the desired prime sum. Define the remainder function $R:(1,\infty)\to\mathbb{C}$ by \begin{align*} R(s):=\sum_p\sum_{k=2}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} The previous step showed that \begin{align*} \log L(s,\chi) =\sum_p \sum_{k=1}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} For each prime $p$, the term with $k=1$ is \begin{align*} \frac{\chi(p)^1}{1\cdot p^s}=\chi(p)p^{-s}. \end{align*} Separating this $k=1$ contribution from all terms with $k\geq 2$ gives \begin{align*} \log L(s,\chi) =\sum_p \chi(p)p^{-s}+R(s). \end{align*} Therefore the theorem will follow once we prove that $R(s)$ is bounded independently of $s$ for all $s$ sufficiently close to $1$ from the right. [/guided] [/step] [step:Bound the higher prime-power contribution uniformly near $s=1$] Let \begin{align*} C_0:=2\sum_{n=2}^{\infty}n^{-2}. \end{align*} The series defining $C_0$ converges, so $C_0<\infty$. For every $1<s\leq 2$, \begin{align*} |R(s)| &\leq \sum_p\sum_{k=2}^{\infty}\frac{|\chi(p)|^k}{k p^{ks}} \\ &\leq \sum_p\sum_{k=2}^{\infty}p^{-ks} \\ &= \sum_p \frac{p^{-2s}}{1-p^{-s}}. \end{align*} Since $p\geq 2$ and $s>1$, we have $p^{-s}\leq 2^{-s}<1/2$, and hence \begin{align*} \frac{1}{1-p^{-s}}\leq 2. \end{align*} Also $p^{-2s}\leq p^{-2}$. Therefore \begin{align*} |R(s)| \leq 2\sum_p p^{-2} \leq 2\sum_{n=2}^{\infty}n^{-2} =C_0. \end{align*} Thus $R(s)=O(1)$ uniformly for $1<s\leq 2$. [guided] We now prove the boundedness of the error term with an explicit constant. Define \begin{align*} C_0:=2\sum_{n=2}^{\infty}n^{-2}. \end{align*} The numerical series $\sum_{n=2}^{\infty}n^{-2}$ converges, so $C_0$ is a finite positive constant. Let $s$ satisfy $1<s\leq 2$. Starting from the definition \begin{align*} R(s):=\sum_p\sum_{k=2}^{\infty}\frac{\chi(p)^k}{k p^{ks}}, \end{align*} we estimate its absolute value by the triangle inequality: \begin{align*} |R(s)| &\leq \sum_p\sum_{k=2}^{\infty}\left|\frac{\chi(p)^k}{k p^{ks}}\right| \\ &= \sum_p\sum_{k=2}^{\infty}\frac{|\chi(p)|^k}{k p^{ks}}. \end{align*} Because $|\chi(p)|\leq 1$ and $1/k\leq 1$ for every integer $k\geq 2$, each summand is bounded by $p^{-ks}$. Hence \begin{align*} |R(s)| &\leq \sum_p\sum_{k=2}^{\infty}p^{-ks}. \end{align*} For each fixed prime $p$, the inner sum is a geometric series with first term $p^{-2s}$ and ratio $p^{-s}$, so \begin{align*} \sum_{k=2}^{\infty}p^{-ks} = \frac{p^{-2s}}{1-p^{-s}}. \end{align*} Now use only the facts that $p\geq 2$ and $s>1$. These imply \begin{align*} p^{-s}\leq 2^{-s}<\frac{1}{2}, \end{align*} so \begin{align*} \frac{1}{1-p^{-s}}\leq 2. \end{align*} Also $s>1$ implies $2s>2$, and therefore \begin{align*} p^{-2s}\leq p^{-2}. \end{align*} Combining these estimates gives \begin{align*} |R(s)| \leq \sum_p \frac{p^{-2s}}{1-p^{-s}} \leq 2\sum_p p^{-2}. \end{align*} Finally, since the set of rational primes is contained in $\{2,3,4,\dots\}$, we have \begin{align*} \sum_p p^{-2}\leq \sum_{n=2}^{\infty}n^{-2}. \end{align*} Therefore \begin{align*} |R(s)|\leq 2\sum_{n=2}^{\infty}n^{-2}=C_0. \end{align*} This bound does not depend on $s$, so the higher prime-power contribution is uniformly bounded as $s\to 1^+$. [/guided] [/step] [step:Conclude the logarithmic prime expansion] For every $1<s\leq 2$, the decomposition \begin{align*} \log L(s,\chi)=\sum_p \chi(p)p^{-s}+R(s) \end{align*} holds, and the preceding step gives $|R(s)|\leq C_0$. Taking $C:=C_0$, where \begin{align*} C_0=2\sum_{n=2}^{\infty}n^{-2}, \end{align*} we have $|R(s)|\leq C$ for all $1<s\leq 2$. Therefore \begin{align*} \log L(s,\chi)=\sum_p \chi(p)p^{-s}+O(1) \end{align*} as $s\to 1^+$. This is the desired expansion. [guided] We combine the two pieces already proved. The separation step gave, for every $s>1$, \begin{align*} \log L(s,\chi)=\sum_p \chi(p)p^{-s}+R(s), \end{align*} where \begin{align*} R(s):=\sum_p\sum_{k=2}^{\infty}\frac{\chi(p)^k}{k p^{ks}}. \end{align*} The uniform bound step proved that, for every $1<s\leq 2$, \begin{align*} |R(s)|\leq C_0, \end{align*} with the explicit finite constant \begin{align*} C_0=2\sum_{n=2}^{\infty}n^{-2}. \end{align*} Thus the constant in the $O(1)$ term may be chosen to be $C:=C_0$. Since this bound is independent of $s$ on the interval $1<s\leq 2$, the remainder stays bounded as $s\to 1^+$. Hence \begin{align*} \log L(s,\chi)=\sum_p \chi(p)p^{-s}+O(1) \end{align*} as $s\to 1^+$, which is exactly the logarithmic prime expansion claimed in the theorem. [/guided] [/step]