# Dirichlet's Theorem on Arithmetic Sequences/Lemma 1

## Lemma for Dirichlet's Theorem on Arithmetic Sequences

Let $a, q$ be coprime integers.

Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.

Let $\chi$ be a Dirichlet character modulo $q$.

Let:

$\displaystyle \map f s = \sum_p \map \chi p p^{-s}$

If $\chi$ is non-trivial then $\map f s$ is bounded as $s \to 1$.

If $\chi$ is the trivial character then:

$\map f s \sim \map \ln {\dfrac 1 {s - 1} }$

as $s \to 1$.

## Proof

$(1):\quad \displaystyle \sum_p \map \chi p p^{-s} = \ln \map L {s, \chi} - \sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} }$

If $\chi$ is non-trivial, then by L-Function does not Vanish at One, $\ln \map L {s, \chi}$ is bounded as $s \to 1$.

If $\chi$ is trivial, then by Analytic Continuation of Dirichlet L-Function, $\map L {s, \chi}$ has a simple pole at $s = 1$.

Therefore, in this case:

$\map L {s, \chi} \sim \dfrac \lambda {s - 1}$

where $\lambda$ is the residue of $\map L {s, \chi}$ at $1$, and:

$\ln \map L {s, \chi} \sim \map \ln {\dfrac \lambda {s - 1} } \sim \map \ln {\dfrac 1 {s - 1} }$

Thus if we can show that the second term of $(1)$ is bounded, the result holds.

On $\map \Re s > 1$:

 $\displaystyle \size {\sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} } }$ $\le$ $\displaystyle \sum_p \sum_{n \mathop \ge 2} \frac 1 {\size {p^s}^n}$ $\displaystyle$ $=$ $\displaystyle \sum_p \frac 1 {\size {p^{2 s} } \, \size {p^s - 1}^n}$ Sum of Infinite Geometric Sequence $\displaystyle$ $\le$ $\displaystyle \sum_p \frac 1 {p^2}$ because $\map \Re s > 1$ $\displaystyle$ $\le$ $\displaystyle \sum_n \frac 1 {n^2}$

This last is $\map \zeta 2$ where $\zeta$ is the Riemann zeta function, so is finite.

$\blacksquare$