(Topics in Complex Analysis I) Convergent Properties of Dirichlet Series

We will take advantage of the convergent properties of Dirichlet series to point out the analytic properties of Riemann zeta function, instead of absolute convergent as we have considered in the previous note.

1. Introduction.

I come back today for the topic of Dirichlet series, and its analytic properties. For the main purpose, we will prove the infiniteness of prime via the main theorem

Theorem 1.1. Let $\zeta(s)=1+2^{-s}+...+3^{-s}$ be the zeta function with complex variable $s$. Then $\zeta(s)$ is meromorphic in the half complex plane $\text{Re } s> 0$ with simple pole at $s=1$, and the residue at $s=1$ is $1$. 

Using this, we can prove

Corollary 1.2. There exists infinite many primes.

Proof. Look back the simple note about Dirichlet's series, for $\text{Re }s>1$one can factorize $\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}$, due to the absolute convergence. By Theorem 1, the simple pole at $s=1$, and the $\text{Res }_{s=1}\zeta(s) = 1$ implies that $\lim_{s\rightarrow1}(s-1)\zeta(s)=1$. Let's take the limit when $\text{Re } s\rightarrow1^{+}$ to obtain

$$1 = \lim_{s\rightarrow1+}(s-1) \prod_{p}\frac{1}{1-p^{-s}}$$

If there are finitely many primes $p$, then the right hand side will be $0$, and that leads the contradiction. That implies there exists infinitely many primes. (Q.E.D)

Via the corollary, we can see an example about how the analytic properties of zeta function play a role in number theory. More than that, one can extend the method of the proof to obtain the same conclusion for Dedekind's zeta function of number fields. To do this, we need study some analytic information of Dirichlet's series, which is the main purpose of the post.

2. Convergent of Dirichlet series.

We first recall the definition of a Dirichlet series. For convention, we always denote $\sigma$ the real part of the complex variable $s$.

Definition 2.1. Let $\{a_n\}$ be a sequence of complex numbers, the Dirichlet series respect to $\{a_n\}$ is $\sum_{n\ge1}\frac{1}{n^s}$.

It can be seen that if $f(s) = \sum_{n\ge1}\frac{1}{n^s}$ absolutely converges for $\text{Re }s = \sigma_0$, then it absolutely converges for $\text{Re }s>\sigma_0$, for $|n^{s}|=n^{\text{Re }s}$. Actually, we also have the same conclusion for convergent information. Let's first prove the lemma.

Lemma 2.2. Assume that for a real number $\sigma_0$, we have $|\sum_{n\le x}\frac{a_n}{n^{\sigma_0}}|\le M$, for all real numbers $x$, $a,b$, and complex number $s$, s.t. $\sigma > \sigma_0$ we have

$$|\sum_{a<n\le b}\frac{a_n}{n^s}| \le Ma^{\sigma_0-\sigma}(1+\frac{|s-s_0|}{\sigma-\sigma_0})$$

Proof. Using the Abel Identity.

Applying the lemma above, we have an important remark.

Remark 2.3. If $|\sum_{n\le x}a_n|$ is bounded by $M$, then we can choose $\sigma_0=0$, and for $s$ s.t. $\sigma>0$, we have $|\sum_{a<n\le b}\frac{a_n}{n^s}|\le 2Ma^{-\sigma}$. And $f(s)$ converges for $s>0$.

For example, we can see that the following Dirichlet's series $\zeta_2(s)=1-2^{-s}+3^{-s}-4^{-s}+...$ is convergent for $\sigma>0$, and so is $\zeta_3(s) = 1 + \frac{1}{2^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}-\frac{2}{6^s}+...$. They are actually play an important role in the proof of Theorem 1.1 above.

We now stay away the very important information about holomorphicity of Dirichlet's series two lemmas.

Lemma 2.4. If the sequence of function $\{f_n(s)\}$ is holomorphic in a region $G$ for each $n$, and for each compact subset of $G$, $f(s) = \lim \{f_n(s)\}$ is uniformly convergent. Then $f(s)$ is also holomorphic in $G$, and $\lim f'_n(s) = f'(s)$.

Proof. Using the complex integral and Cauchy's theorem, with the note that when the sequence of function is absolutely convergent, we can interchange the limit and the integral.

Lemma 2.5. Let $f(s)=\sum_{n\ge 1}\frac{a_n}{n^s}$ converges for $\sigma > \sigma_c$ then $f(s)$ is uniformly convergent in each compact subset of the half plane $\sigma>\sigma_c$.

Proof. We just need to consider the rectangle that contains $s$, and then apply Lemma 2.2.

We now come to the very important

Theorem 2.6. Let $f(s)=\sum_{n\ge 1}\frac{a_n}{n^s}$ be a Dirichlet series, that converges for $\sigma>\sigma_c$ then it is holomorphic in the half plane $\sigma>\sigma_c$.

Proof. It is just the combination of Lemma 2.4 and Lemma 2.5.

Corollary. If $|\sum_{n\le x}a_n|$ is bounded for all $x$, then $f(s)$ is holomorphic in the half plane $\sigma>0$. As a consequence, $\zeta_2(s),\zeta_3(s)$ are holomorphic in the half plane $\sigma>0$.

We are now ready for the

Proof of Theorem 1.1. It can be seen that for $\sigma>1$, we have $\zeta_2(s)+2^{1-s}\zeta(s)=\zeta(s)$. This implies $\zeta_2(s)=(1-2^{1-s})\zeta(s)$. Hence, $\zeta(s)$ is the meromorphic function in the half plane $\sigma>0$, with simple poles at the zeros of $1-2^{1-s}$. In this case $s\in\{1-2n\pi i/\log 2|n\in \mathbb{Z}\}$. Also, $\zeta_3(s)=(1-3^{1-s})\zeta(s)$, and it is meromorphic function in the same half plane as above with poles in the set $\{1-2m\pi i/\log 3|m\in \mathbb{Z}\}$. This fact implies that $s=1$ is the only simple pole (for $2^n=3^m$ implies that m = n = 0).

For the residue, we just need to compute $\lim_{s\rightarrow 1}(s-1)\zeta(s)$. By applying the integral inequality in integral test, we can see

$$\int_1^\infty x^{-s}dx\le\zeta(s)\le1 +\int_1^\infty x^{-s}dx$$

Let $y=x^{-1}$, then $\int_1^\infty x^{-s}dx = \int_0^1 y^{s-2}dy = \frac{1}{s-1}$. By simple calculation, one can see $1\le (s-1)\zeta(s) \le s$. That implies the residue of $\zeta(s)$ at $s_0 = 1$. (Q.E.D)