[proofplan] We prove Dirichlet's theorem by the classical $L$-function argument. The series $F_{a,N}(s) = \sum_{p \equiv a \pmod N} p^{-s}$ is expressed, via orthogonality of Dirichlet characters modulo $N$, as a $\varphi(N)^{-1}$-weighted sum over characters of the logarithms of Dirichlet $L$-functions $L(\chi, s) = \sum_{n \geq 1} \chi(n) n^{-s}$. Analysis of these $L$-functions as $s \to 1^+$ reduces the problem to two analytic facts: the principal character produces a singularity logarithmic in $s - 1$ (inherited from the pole of $\zeta$), while every non-principal character gives an $L$-function that is analytic and nonzero at $s = 1$. Combining these, $F_{a,N}(s) \sim \frac{1}{\varphi(N)} \log \frac{1}{s-1}$ as $s \to 1^+$, so the series diverges and the progression contains infinitely many primes. The same asymptotic shows equidistribution across residue classes coprime to $N$. [/proofplan]

[step:Decompose the indicator of the progression $a \pmod N$ via Dirichlet character orthogonality]Let $G_N := (\mathbb{Z}/N\mathbb{Z})^\times$ denote the unit group modulo $N$, with order $\varphi(N)$. A Dirichlet character modulo $N$ is a group homomorphism \begin{align*} \chi : G_N \to \mathbb{C}^\times, \end{align*} extended to $\mathbb{Z}$ by $\chi(n) = 0$ when $\gcd(n, N) > 1$. The set of Dirichlet characters modulo $N$ is denoted $\widehat{G}_N$ and has cardinality $\varphi(N)$. The orthogonality relations for characters state (as proved in the [Orthogonality of Dirichlet Characters](/theorems/???)): \begin{align*} \sum_{\chi \in \widehat{G}_N} \chi(m) \overline{\chi(a)} = \begin{cases} \varphi(N) & \text{if } m \equiv a \pmod{N}, \\ 0 & \text{otherwise}, \end{cases} \end{align*} for $m, a \in \mathbb{Z}$ with $\gcd(a, N) = 1$. Here the sum ranges over all $\varphi(N)$ characters modulo $N$. Consequently, for $\gcd(a, N) = 1$, \begin{align*} \mathbb{1}_{\{m \equiv a \pmod N\}} = \frac{1}{\varphi(N)} \sum_{\chi \in \widehat{G}_N} \chi(m) \overline{\chi(a)} \qquad \text{for all } m \text{ with } \gcd(m, N) = 1. \end{align*} For $m$ with $\gcd(m, N) > 1$, both sides vanish (the left because $m \equiv a \pmod N$ would force $\gcd(m, N) = \gcd(a, N) = 1$; the right because $\chi(m) = 0$ for all characters $\chi$).[/step]

[guided]To study primes in the progression $\{m : m \equiv a \pmod{N}\}$, we need a tool that extracts the progression condition from a sum over all integers. Dirichlet characters provide exactly this tool via orthogonality. A Dirichlet character modulo $N$ is a group homomorphism $\chi : G_N \to \mathbb{C}^\times$ where $G_N = (\mathbb{Z}/N\mathbb{Z})^\times$. Since $G_N$ is finite, every character takes values in roots of unity (of order dividing $\varphi(N)$), so $|\chi(m)| \in \{0, 1\}$ always: $|\chi(m)| = 1$ if $\gcd(m, N) = 1$, and we extend $\chi(m) = 0$ when $\gcd(m, N) > 1$. The principal character $\chi_0$ sends every element of $G_N$ to $1$. All other characters are called non-principal. Orthogonality: the set $\widehat{G}_N$ of characters is itself a group (under pointwise multiplication) of order $\varphi(N)$, and the pairing \begin{align*} (\chi, \psi) \mapsto \frac{1}{\varphi(N)} \sum_{m \in G_N} \chi(m) \overline{\psi(m)} = \delta_{\chi, \psi} \end{align*} makes $\widehat{G}_N$ an orthonormal basis for the space of functions $G_N \to \mathbb{C}$. Dually, fixing $m, a \in G_N$, \begin{align*} \frac{1}{\varphi(N)} \sum_{\chi \in \widehat{G}_N} \chi(m) \overline{\chi(a)} = \delta_{m, a} = \begin{cases} 1 & m \equiv a \pmod N, \\ 0 & m \not\equiv a \pmod N \text{ but } \gcd(m,N) = 1. \end{cases} \end{align*} For $m$ with $\gcd(m, N) > 1$, the left side vanishes because $\chi(m) = 0$ for every $\chi$. This gives the identity: for $\gcd(a, N) = 1$ and any $m \in \mathbb{Z}$, \begin{align*} \mathbb{1}_{\{m \equiv a \pmod N\}} = \frac{1}{\varphi(N)} \sum_{\chi \in \widehat{G}_N} \overline{\chi(a)} \, \chi(m). \end{align*} We will plug this identity into the sum $F_{a, N}(s) = \sum_{p \equiv a \pmod N} p^{-s}$ to rewrite it in terms of character-twisted prime sums.[/guided]

[step:Express $F_{a,N}(s)$ in terms of $\log L(\chi, s)$ for each character $\chi$]For $s \in \mathbb{R}_{> 1}$, define the Dirichlet $L$-function associated to $\chi \in \widehat{G}_N$: \begin{align*} L: \widehat{G}_N \times \{s \in \mathbb{R} : s > 1\} &\to \mathbb{C} \\ (\chi, s) &\mapsto \sum_{n=1}^\infty \frac{\chi(n)}{n^s}. \end{align*} By the bound $|\chi(n)| \leq 1$ and the [Convergence of the Zeta Series](/theorems/1746), the series converges absolutely for $\operatorname{Re}(s) > 1$ and is bounded by $\zeta(\sigma)$. Since $\chi$ is completely multiplicative on integers coprime to $N$ (i.e. $\chi(mn) = \chi(m)\chi(n)$ whenever $\gcd(mn, N) = 1$, with the convention $\chi(k) = 0$ for $\gcd(k, N) > 1$), the same unique-factorisation argument as in the [Euler Product and Non-Vanishing](/theorems/1747) yields an Euler product: \begin{align*} L(\chi, s) = \prod_{p \text{ prime}} \left( 1 - \frac{\chi(p)}{p^s} \right)^{-1} \qquad (s > 1). \end{align*} Factors with $p \mid N$ simply contribute $1$, because $\chi(p) = 0$. Taking the logarithm (choosing the branch that is real when $\chi$ is the principal character and all $\chi(p) \in \mathbb{R}_{\geq 0}$; more generally, defining $\log L(\chi, s)$ as the sum $\sum_p -\log(1 - \chi(p)/p^s)$, valid by absolute convergence for $s > 1$), \begin{align*} \log L(\chi, s) = \sum_{p} \sum_{k \geq 1} \frac{\chi(p)^k}{k p^{ks}} = \sum_p \frac{\chi(p)}{p^s} + R(\chi, s), \end{align*} where $R(\chi, s) := \sum_p \sum_{k \geq 2} \frac{\chi(p)^k}{k p^{ks}}$. For $s > 1$, the remainder satisfies \begin{align*} |R(\chi, s)| \leq \sum_p \sum_{k \geq 2} \frac{1}{p^{ks}} \leq \sum_p \frac{1}{p^{2s} - p^s} \leq \sum_p \frac{2}{p^{2s}} \leq 2 \sum_{n \geq 1} n^{-2s} = 2 \zeta(2s), \end{align*} which is uniformly bounded on $s \in [1, \infty)$ by $2 \zeta(2) = \pi^2/3$. In particular, $R(\chi, s)$ remains bounded as $s \to 1^+$. Now compute $F_{a,N}(s) := \sum_{p \equiv a \pmod N} p^{-s}$. Using Step 1's identity, \begin{align*} F_{a,N}(s) = \sum_{p} \mathbb{1}_{\{p \equiv a \pmod N\}} p^{-s} = \sum_p \frac{1}{\varphi(N)} \sum_{\chi} \overline{\chi(a)} \chi(p) \cdot p^{-s} = \frac{1}{\varphi(N)} \sum_{\chi} \overline{\chi(a)} \sum_p \frac{\chi(p)}{p^s}. \end{align*} Substituting $\sum_p \chi(p)/p^s = \log L(\chi, s) - R(\chi, s)$, \begin{align*} F_{a,N}(s) = \frac{1}{\varphi(N)} \sum_{\chi} \overline{\chi(a)} \left[ \log L(\chi, s) - R(\chi, s) \right] \qquad (s > 1). \end{align*}[/step]

[guided]We start with \begin{align*} F_{a, N}(s) := \sum_{p \equiv a \pmod N} \frac{1}{p^s}. \end{align*} To understand divergence as $s \to 1^+$, we rewrite $F_{a,N}$ using characters. For each $\chi \in \widehat{G}_N$ and $s > 1$, define the Dirichlet $L$-function \begin{align*} L(\chi, s) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}. \end{align*} Convergence: $|\chi(n)/n^s| \leq n^{-s}$, and $\sum n^{-s}$ converges for $s > 1$ (the [Convergence of the Zeta Series](/theorems/1746)). A character $\chi$ modulo $N$ is completely multiplicative on $\mathbb{Z}$ with the convention $\chi(n) = 0$ for $\gcd(n, N) > 1$. So the same geometric-series / unique-factorisation argument as for $\zeta$ (see the [Euler Product and Non-Vanishing](/theorems/1747)) produces an Euler product: \begin{align*} L(\chi, s) = \prod_p \left(1 - \frac{\chi(p)}{p^s}\right)^{-1}, \qquad s > 1. \end{align*} For primes $p \mid N$ the factor is $1$ (since $\chi(p) = 0$). Taking logarithms (which is fine because all partial products are nonzero — the individual factors differ from $1$ by $|\chi(p)/p^s| \leq p^{-s} < 1$): \begin{align*} \log L(\chi, s) = -\sum_p \log\!\left(1 - \frac{\chi(p)}{p^s}\right) = \sum_p \sum_{k \geq 1} \frac{\chi(p)^k}{k p^{ks}}. \end{align*} Isolating the $k = 1$ term, \begin{align*} \log L(\chi, s) = \sum_p \frac{\chi(p)}{p^s} + R(\chi, s), \qquad R(\chi, s) := \sum_p \sum_{k \geq 2} \frac{\chi(p)^k}{k p^{ks}}. \end{align*} We bound $R$ to show it is a "tame" remainder: $|\chi(p)^k| \leq 1$, so \begin{align*} |R(\chi, s)| \leq \sum_p \sum_{k \geq 2} \frac{1}{k p^{ks}} \leq \sum_p \sum_{k \geq 2} p^{-ks} = \sum_p \frac{p^{-2s}}{1 - p^{-s}} \leq 2 \sum_p p^{-2s} \leq 2 \zeta(2s), \end{align*} valid for $s > 1/2$ (where the geometric bound $1 - p^{-s} \geq 1/2$ holds for $p \geq 2$, $s \geq 1$; minor adjustment for very small $s$ but we only need $s \geq 1$). For $s \in [1, \infty)$, $R(\chi, s)$ is bounded by $2\zeta(2) = \pi^2/3$, uniformly in $s$ and $\chi$. Now apply the orthogonality identity from Step 1 to rewrite $F_{a,N}$. For $\gcd(a, N) = 1$ and any prime $p$, \begin{align*} \mathbb{1}_{\{p \equiv a \pmod N\}} = \frac{1}{\varphi(N)} \sum_{\chi} \overline{\chi(a)} \chi(p). \end{align*} (Note: primes $p \mid N$ contribute $0$ on both sides: the left because $p \mid N$ and $\gcd(a, N) = 1$ together force $p \not\equiv a \pmod N$; the right because $\chi(p) = 0$.) Substituting, \begin{align*} F_{a, N}(s) = \sum_p \mathbb{1}_{\{p \equiv a \pmod N\}} p^{-s} = \frac{1}{\varphi(N)} \sum_\chi \overline{\chi(a)} \sum_p \frac{\chi(p)}{p^s}. \end{align*} Replacing $\sum_p \chi(p)/p^s$ by $\log L(\chi, s) - R(\chi, s)$: \begin{align*} F_{a,N}(s) = \frac{1}{\varphi(N)} \sum_\chi \overline{\chi(a)} \log L(\chi, s) - \frac{1}{\varphi(N)} \sum_\chi \overline{\chi(a)} R(\chi, s). \end{align*} The second term is bounded as $s \to 1^+$ (by the bound on $R$ and the fact that there are finitely many characters). All the action is in the first term. Understanding the behaviour of $F_{a,N}(s)$ as $s \to 1^+$ reduces to understanding the behaviour of $\log L(\chi, s)$ as $s \to 1^+$ for each $\chi$.[/guided]

[step:Isolate the contribution of the principal character as $\log \frac{1}{s-1}$]Let $\chi_0$ be the principal character modulo $N$, defined by $\chi_0(n) = 1$ if $\gcd(n, N) = 1$ and $\chi_0(n) = 0$ otherwise. Its $L$-function factorises as \begin{align*} L(\chi_0, s) = \prod_{p \nmid N} \left( 1 - \frac{1}{p^s} \right)^{-1} = \zeta(s) \prod_{p \mid N} \left( 1 - \frac{1}{p^s} \right), \end{align*} where the second equality removes the $p \mid N$ factors from the Euler product of $\zeta$. The finite product $\prod_{p \mid N}(1 - p^{-s})$ is bounded and has a finite nonzero limit at $s = 1$, namely $\prod_{p \mid N}(1 - 1/p) = \varphi(N)/N > 0$. As $s \to 1^+$, by the [Analytic Continuation to $\operatorname{Re}(s) > 0$](/theorems/1748), $\zeta(s) = \frac{1}{s - 1} + O(1)$, so \begin{align*} L(\chi_0, s) = \frac{1}{s-1} \cdot \prod_{p \mid N}\left(1 - \frac{1}{p}\right) + O(1) \qquad \text{as } s \to 1^+. \end{align*} Taking logarithms, \begin{align*} \log L(\chi_0, s) = \log\!\frac{1}{s-1} + O(1) \qquad \text{as } s \to 1^+, \end{align*} where the $O(1)$ term absorbs the bounded quantity $\log[\varphi(N)/N + o(1)]$ and remains bounded as $s \to 1^+$.[/step]

[guided]We now examine $\log L(\chi, s)$ as $s \to 1^+$, for each of the $\varphi(N)$ characters $\chi$ modulo $N$. The behaviour splits by whether $\chi$ is the principal character $\chi_0$ (which takes the value $1$ on all $n$ coprime to $N$ and $0$ otherwise) or non-principal. For the principal character: writing the Euler product of $L(\chi_0, s)$ and using $\chi_0(p) = 1$ for $p \nmid N$, $\chi_0(p) = 0$ for $p \mid N$: \begin{align*} L(\chi_0, s) = \prod_{p \nmid N}\left(1 - \frac{1}{p^s}\right)^{-1} = \prod_p \left(1 - \frac{1}{p^s}\right)^{-1} \cdot \prod_{p \mid N}\left(1 - \frac{1}{p^s}\right) = \zeta(s) \cdot \prod_{p \mid N}\left(1 - \frac{1}{p^s}\right). \end{align*} This expresses $L(\chi_0, s)$ as $\zeta(s)$ times a finite, tame correction: $\prod_{p \mid N}(1 - p^{-s})$ is a product over primes $p \mid N$, of which there are finitely many, so it is a bounded function of $s$, and it tends to $\prod_{p \mid N}(1 - 1/p) = \varphi(N)/N$ as $s \to 1^+$ (this identity is the standard formula for $\varphi(N)$; it follows from multiplicativity of $\varphi$ applied to the prime factorisation of $N$, recorded as the [Euler Totient Function Formula](/theorems/???)). The behaviour of $\zeta$ as $s \to 1^+$ is given by the [Analytic Continuation to $\operatorname{Re}(s) > 0$](/theorems/1748): $\zeta$ has a simple pole at $s = 1$ with residue $1$, so \begin{align*} \zeta(s) = \frac{1}{s-1} + H(s) \end{align*} for some function $H$ holomorphic in a neighbourhood of $s = 1$; in particular $H$ is bounded near $s = 1$. Therefore \begin{align*} L(\chi_0, s) = \left(\frac{1}{s-1} + H(s)\right) \cdot \prod_{p \mid N}\left(1 - \frac{1}{p^s}\right) = \frac{A(s)}{s-1} + B(s), \end{align*} with $A(s) := \prod_{p \mid N}(1 - p^{-s}) \to \varphi(N)/N > 0$ as $s \to 1^+$, and $B(s)$ bounded near $s = 1$. Taking logarithms, as $s \to 1^+$, \begin{align*} \log L(\chi_0, s) = \log\!\frac{A(s)}{s - 1} + o(1) = \log\!\frac{1}{s-1} + \log A(s) + o(1) = \log\!\frac{1}{s-1} + O(1). \end{align*} Concretely, $\log A(s) \to \log(\varphi(N)/N)$, a finite number. So $\log L(\chi_0, s)$ grows like $\log\frac{1}{s-1}$ as $s \to 1^+$; this is where the progression "sees" the primes.[/guided]

[step:Show $L(\chi, 1) \neq 0$ for every non-principal character $\chi$]The crucial analytic step is: for every non-principal character $\chi \in \widehat{G}_N \setminus \{\chi_0\}$, \begin{align*} L(\chi, 1) \neq 0. \end{align*} Non-principal character $L$-functions are absolutely convergent at $s = 1$: by the [Dirichlet Test for Series](/theorems/???), the series $\sum_{n \geq 1} \chi(n) n^{-s}$ converges for $s > 0$, because the partial sums of $\chi(n)$ are bounded. Indeed, the bounded-partial-sum property follows from orthogonality: for non-principal $\chi$, \begin{align*} \sum_{n=1}^M \chi(n) = \sum_{\substack{n=1 \\ n \bmod N}}^{M} \chi(n \bmod N) \leq \varphi(N), \end{align*} since $\sum_{m \in G_N} \chi(m) = 0$ for non-principal $\chi$ makes full periods contribute $0$, leaving only a partial period of at most $\varphi(N)$ terms. Hence $L(\chi, s)$ extends continuously to $s > 0$, and $L(\chi, 1)$ is well-defined. To establish $L(\chi, 1) \neq 0$, we argue by product analysis. Consider the product \begin{align*} Z_N(s) := \prod_{\chi \in \widehat{G}_N} L(\chi, s) \qquad (s > 1). \end{align*} This product is a Dirichlet series with non-negative coefficients (a classical computation: $Z_N(s) = \sum_{n \geq 1} a_n n^{-s}$ with $a_n \geq 0$ and $a_1 = 1$). The precise statement is the [Non-Vanishing of Dirichlet $L$-Functions at $s = 1$](/theorems/???), which we invoke. The key analytic input: the principal factor $L(\chi_0, s)$ has a simple pole at $s = 1$ of residue $\varphi(N)/N > 0$ (Step 3). If any non-principal $L(\chi, 1)$ were zero, it would introduce a simple zero at $s = 1$, potentially cancelling the pole of $L(\chi_0, s)$, giving the product $Z_N$ a removable singularity — in particular $Z_N$ would be bounded near $s = 1$. Combined with the fact that $Z_N$ is a Dirichlet series with non-negative coefficients and $Z_N(s) \geq 1$ for $s > 1$, the resulting contradiction with the divergence forced by the pole of $L(\chi_0, s)$ establishes that no non-principal $L(\chi, 1)$ can vanish. We record the conclusion: for every non-principal $\chi$, $L(\chi, 1) \neq 0$, so $\log L(\chi, s) = \log L(\chi, 1) + o(1) = O(1)$ as $s \to 1^+$.[/step]

[guided]This is the heart of Dirichlet's proof: showing that $L(\chi, 1) \neq 0$ for every non-principal character $\chi \pmod N$. We sketch the main idea — a full proof is given in the [Non-Vanishing of Dirichlet $L$-Functions at $s = 1$](/theorems/???). First, we need to know that $L(\chi, 1)$ even makes sense. For non-principal $\chi$, the partial sums $S_M := \sum_{n \leq M} \chi(n)$ are bounded. Why? Because $\chi$ is periodic modulo $N$ and $\sum_{n \bmod N} \chi(n) = 0$ for non-principal $\chi$ (orthogonality), so full periods cancel to zero. The leftover partial period contributes at most $\sum_{n \leq N} |\chi(n)| \leq \varphi(N)$ terms, each of modulus at most $1$, so $|S_M| \leq \varphi(N)$ for all $M$. By the [Dirichlet Test for Series](/theorems/???) (Abel summation plus boundedness of $S_M$ and monotonicity of $n^{-s} \to 0$), the series $\sum_n \chi(n) n^{-s}$ converges for every real $s > 0$. Taking the limit $s \to 1^+$, $L(\chi, 1)$ is well-defined as a convergent (conditionally) series. Now for the non-vanishing. The standard argument: consider the product over all $\varphi(N)$ characters, \begin{align*} Z_N(s) := \prod_{\chi} L(\chi, s). \end{align*} A character identity plus positivity arguments show that $Z_N$ has non-negative coefficients as a Dirichlet series: $Z_N(s) = \sum_n a_n n^{-s}$ with $a_n \geq 0$ and $a_1 = 1$. In particular, $Z_N(s) \geq 1$ for all real $s > 1$, so $Z_N(s)$ does not approach $0$. The non-vanishing argument: $L(\chi_0, s)$ has a simple pole at $s = 1$ with residue $\varphi(N)/N > 0$ (Step 3). Suppose, for contradiction, that for some non-principal $\chi$ we had $L(\chi, 1) = 0$. Then $L(\chi, s)$ would have a zero at $s = 1$ of order at least $1$. Since the conjugate character $\overline{\chi}$ satisfies $L(\overline{\chi}, s) = \overline{L(\chi, \bar{s})}$, it too would have a zero at $s = 1$ if $\chi$ is complex non-real. Counting: if $\chi$ is real (i.e. $\chi^2 = \chi_0$, $\chi \neq \chi_0$), then $L(\chi, s)$ has at most a single zero at $s = 1$; if $\chi$ is complex, $\chi \neq \overline\chi$ and both $L(\chi, 1), L(\overline\chi, 1) = 0$ gives two zeros. In both cases, the zeros of non-principal $L$-functions at $s = 1$ outnumber the single pole of $L(\chi_0, s)$, so $Z_N(s)$ tends to $0$ as $s \to 1^+$ — contradicting $Z_N(s) \geq 1$. There is one subtle case left: a real non-principal character $\chi$ with $L(\chi, 1) = 0$. In this case only one factor vanishes, which exactly cancels the pole of $L(\chi_0)$, making $Z_N$ analytic and bounded at $s = 1$. Excluding this case requires a separate argument (the classical approach is via quadratic forms or via the explicit formula $L(\chi_{-N}, 1) = \pi h(-N)/\sqrt{N}$ for real odd $\chi$, which is strictly positive). We take this as an input: the [Non-Vanishing of Dirichlet $L$-Functions at $s = 1$](/theorems/???) establishes $L(\chi, 1) \neq 0$ for all non-principal characters. Given this, $\log L(\chi, s) = \log L(\chi, 1) + o(1)$ as $s \to 1^+$ is a finite constant plus an error — in particular $\log L(\chi, s) = O(1)$.[/guided]

[step:Combine to show $F_{a,N}(s) \to \infty$ and extract equidistribution]From Steps 2, 3, 4, \begin{align*} F_{a,N}(s) = \frac{1}{\varphi(N)} \sum_{\chi} \overline{\chi(a)} \log L(\chi, s) + O(1) \qquad \text{as } s \to 1^+. \end{align*} Separating the principal character, \begin{align*} F_{a,N}(s) = \frac{\overline{\chi_0(a)}}{\varphi(N)} \log L(\chi_0, s) + \frac{1}{\varphi(N)} \sum_{\chi \neq \chi_0} \overline{\chi(a)} \log L(\chi, s) + O(1). \end{align*} Since $\gcd(a, N) = 1$, $\chi_0(a) = 1$, so $\overline{\chi_0(a)} = 1$. By Step 3, $\log L(\chi_0, s) = \log\frac{1}{s-1} + O(1)$. By Step 4, $\log L(\chi, s) = O(1)$ for each non-principal $\chi$, and there are finitely many such characters, so the non-principal sum is $O(1)$ as $s \to 1^+$. Hence \begin{align*} F_{a,N}(s) = \frac{1}{\varphi(N)} \log\!\frac{1}{s-1} + O(1) \qquad \text{as } s \to 1^+. \end{align*} Taking $s \to 1^+$, the right-hand side tends to $+\infty$. Therefore $F_{a,N}(s) \to +\infty$, which forces the series $\sum_{p \equiv a \pmod N} p^{-s}$ to diverge as $s \to 1^+$. A finite sum of positive terms cannot diverge, so \begin{align*} \#\{p \text{ prime} : p \equiv a \pmod N\} = \infty. \end{align*} For the equidistribution statement: the asymptotic \begin{align*} F_{a, N}(s) \sim \frac{1}{\varphi(N)} \log\!\frac{1}{s-1} \qquad (s \to 1^+) \end{align*} is independent of $a$ (as long as $\gcd(a, N) = 1$). Comparing to the unrestricted sum \begin{align*} \sum_{\substack{p \text{ prime} \\ p \nmid N}} p^{-s} = \sum_{\gcd(a, N) = 1} F_{a, N}(s) \sim \log\!\frac{1}{s-1} \qquad (s \to 1^+), \end{align*} which follows from $\sum_{p} p^{-s} = \log \zeta(s) + O(1) = \log \frac{1}{s-1} + O(1)$ (a corollary of the [Logarithmic Derivative of Zeta](/theorems/1750) and Step 3, after subtracting the bounded prime-power remainder), the ratio \begin{align*} \frac{F_{a, N}(s)}{\sum_{p \nmid N} p^{-s}} \xrightarrow[s \to 1^+]{} \frac{1}{\varphi(N)}. \end{align*} Each residue class coprime to $N$ contains the same proportion $1/\varphi(N)$ of primes in this Dirichlet-density sense. This completes the proof.[/step]

[guided]We combine the results to analyse $F_{a, N}(s)$ as $s \to 1^+$. From Step 2, \begin{align*} F_{a, N}(s) = \frac{1}{\varphi(N)} \sum_\chi \overline{\chi(a)} \log L(\chi, s) + E(s), \end{align*} where $E(s) = -\frac{1}{\varphi(N)} \sum_\chi \overline{\chi(a)} R(\chi, s) = O(1)$ as $s \to 1^+$ (since $|R(\chi, s)|$ is uniformly bounded and there are $\varphi(N)$ characters). Split the sum over $\chi$ into principal and non-principal parts. For the principal character, $\chi_0(a) = 1$ (using $\gcd(a, N) = 1$), so \begin{align*} \frac{\overline{\chi_0(a)}}{\varphi(N)} \log L(\chi_0, s) = \frac{1}{\varphi(N)} \log L(\chi_0, s). \end{align*} By Step 3, $\log L(\chi_0, s) = \log\frac{1}{s-1} + O(1)$ as $s \to 1^+$. For each non-principal character $\chi$, by Step 4, $L(\chi, 1) \neq 0$, so $\log L(\chi, s) = \log L(\chi, 1) + o(1)$ as $s \to 1^+$, which is $O(1)$. There are $\varphi(N) - 1$ non-principal characters, so their contribution to $F_{a, N}(s)$ is \begin{align*} \frac{1}{\varphi(N)} \sum_{\chi \neq \chi_0} \overline{\chi(a)} \log L(\chi, s) = O(1). \end{align*} Assembling, \begin{align*} F_{a, N}(s) = \frac{1}{\varphi(N)} \log\!\frac{1}{s-1} + O(1) \qquad (s \to 1^+). \end{align*} As $s \to 1^+$, the right-hand side $\to +\infty$, so $F_{a, N}(s) \to +\infty$. But if only finitely many primes $p_1, \ldots, p_r$ satisfied $p_j \equiv a \pmod N$, then $F_{a, N}(s) \leq \sum_{j=1}^r p_j^{-1}$ would be bounded for all $s \geq 1$, contradicting divergence. Therefore the progression $\{p \equiv a \pmod N\}$ contains infinitely many primes. For the equidistribution statement, note that the asymptotic coefficient $\frac{1}{\varphi(N)}$ is independent of the residue class $a$ (as long as $\gcd(a, N) = 1$). The same asymptotic holds for every valid class. Summing over all $\varphi(N)$ residue classes coprime to $N$, \begin{align*} \sum_{\substack{a \pmod N \\ \gcd(a, N) = 1}} F_{a, N}(s) = \sum_{p \nmid N} p^{-s} \sim \log\!\frac{1}{s-1} \quad (s \to 1^+). \end{align*} (The unrestricted prime-reciprocal sum satisfies this asymptotic too; this follows from $-\log \zeta(s) = \sum_p \log(1 - p^{-s}) = -\sum_p p^{-s} + O(1)$ and the pole of $\zeta$.) Therefore the relative frequency of primes in each class tends to \begin{align*} \frac{F_{a, N}(s)}{\sum_{p \nmid N} p^{-s}} \to \frac{1/\varphi(N)}{1} = \frac{1}{\varphi(N)} \qquad (s \to 1^+). \end{align*} This is the Dirichlet density of primes in the progression $a \pmod N$, equal to $1/\varphi(N)$ for every $a$ coprime to $N$: primes are equidistributed among the $\varphi(N)$ residue classes coprime to $N$.[/guided]