Dirichlet's Theorem on Arithmetic Sequences/Lemma 1
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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:
- $\ds \map f s = \sum_p \map \chi p p^{-s}$
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
By Logarithm of Dirichlet L-Functions:
- $(1): \quad \ds \sum_p \map \chi p p^{-s} = \ln \map L {s, \chi} - \sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} }$
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 $\map \Re s > 1$:
| \(\ds \size {\sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} } }\) | \(\le\) | \(\ds \sum_p \sum_{n \mathop \ge 2} \frac 1 {\size {p^s}^n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_p \frac 1 {\size {p^{2 s} } \, \size {p^s - 1}^n}\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \sum_p \frac 1 {p^2}\) | because $\map \Re s > 1$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \sum_n \frac 1 {n^2}\) |
This last is $\map \zeta 2$ where $\zeta$ is the Riemann zeta function, so is finite.
$\blacksquare$