Let $\chi$ be a non-principal Dirichlet character modulo $n$. We prove $L(1, \chi) \neq 0$.

Step 1: Logarithmic derivative

For $\text{Re}(s) > 1$: $-\frac{L'(s, \chi)}{L(s, \chi)} = \sum_p \frac{\chi(p)\log p}{p^s} + O(1)$

This follows from the Euler product $L(s, \chi) = \prod_p (1 - \chi(p)p^{-s})^{-1}$.

Step 2: Consider the product over all characters

For characters $\chi_0, \chi_1, \ldots, \chi_{\varphi(n)-1}$ modulo $n$: $\prod_{j=0}^{\varphi(n)-1}L(s, \chi_j) = \zeta(s) \cdot \prod_{j=1}^{\varphi(n)-1}L(s, \chi_j)$

where $\chi_0$ is the principal character. Taking logarithmic derivatives: $\sum_{j=0}^{\varphi(n)-1}\frac{L'(s, \chi_j)}{L(s, \chi_j)} = \sum_p \frac{A_p\log p}{p^s}$

where $A_p = \sum_j \chi_j(p)$ depends on $p \bmod n$.

Step 3: Orthogonality of characters

By character orthogonality: $\sum_j \chi_j(a) = \varphi(n)$ if $a \equiv 1 \pmod{n}$ and $0$ otherwise.

Therefore, the sum $-\frac{L'(s, \chi_0)}{L(s, \chi_0)}$ has a positive contribution from all primes, dominating other terms.

Step 4: Approaching $s = 1$

As $s \to 1^+$: $-\frac{L'(s, \chi_0)}{L(s, \chi_0)} = \frac{1}{s-1} + O(1)$

since $L(s, \chi_0)$ has a simple pole at $s = 1$ (essentially $\zeta(s)$ with finitely many Euler factors removed).

Step 5: Contradiction if $L(1, \chi) = 0$

If $L(1, \chi_j) = 0$ for some non-principal $\chi_j$, then $-\frac{L'(s, \chi_j)}{L(s, \chi_j)}$ has a simple pole at $s = 1$ with residue related to the order of vanishing.

Summing over all characters, the left side becomes unbounded, but the right side: $\sum_p \frac{A_p\log p}{p^s}$

is bounded as $s \to 1^+$ by character orthogonality (only finitely many primes contribute per residue class).

This contradiction proves $L(1, \chi) \neq 0$ for all non-principal $\chi$.

Step 6: Implication for primes

The non-vanishing implies: $\sum_{p \equiv a \pmod{n}}\frac{1}{p} = \infty$

Therefore infinitely many primes $p \equiv a \pmod{n}$ exist, with asymptotic density $1/\varphi(n)$ following from more refined analysis.