Difference between revisions of "Prime Number Theorem"

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Introduction

The aim of this article is to present the proof of the prime number theorem that says that the number $\pi(n)$ of primes not exceeding $n$ is approximately $\frac n{\ln n}$ or, more precisely, that

$\lim_{n\to\infty}\frac {\pi(n)\ln n}{n}=1.$

Unfortunately, all known proofs of the prime number theorem are quite involved and require knowledge of some college level material.

We shall start with some elementary observations due to Chebyshev.

Chebyshev's estimates

The key observation is that ${2m\choose m}=\frac{(2m)!}{(m!)^2}$ is divisible by all primes $p$ satisfying $m<p\le 2m$ . Indeed, every such prime appears in the numerator $(2m)!$ and does not appear in the denominator $(m!)^2$ . Thus, $\prod_{m<p\le 2m}p\le {2m\choose m}\le 2^{2m}$ (from now on, $p$ will always stand for a prime number, so, to shorten the notation, we will not explicitly write " $p$ is prime" in the descriptions of sum and product ranges). Similarly, considering ${2m+1\choose m}=\frac{(2m+1)!}{m!(m+1)!}$ , we see that $\prod_{m+1<p\le 2m+1}p\le {2m+1\choose m}\le 2^{2m}$ (the last inequality holds because ${2m+1\choose m}={2m+1\choose m+1}$ and ${2m+1\choose m}+{2m+1\choose m+1}\le 2^{2m+1}$ ). Now we are ready to prove the

Upper bound for the product of primes

Lemma 1

For every positive integer $n$ , we have $\prod_{p\le n}p\le 4^n$

Proof of Lemma 1

Induction on $n$ . The base $n=1$ is obvious. Suppose now that the statement holds for all numbers strictly less than $n$ . If $n=2m$ is even, then

$\prod_{p\le n}p=\left(\prod_{p\le m}p\right)\cdot \left(\prod_{m<p\le 2m}p\right)\le 4^m 2^{2m}=4^{2m}=4^n.$

If $n=2m+1$ is odd, then

$\prod_{p\le n}p=\left(\prod_{p\le m+1}p\right)\cdot \left(\prod_{m+1<p\le 2m+1}p\right)\le 4^{m+1} 2^{2m}=4^{2m+1}=4^n.$

From this lemma we can easily derive the

Upper bound for the number of primes

Lemma 2

$\pi(n)\le 4\frac n{\ln n}$ for large $n$

Proof of Lemma 2

Rewrite the inequality of Lemma 1 as $\sum_{p\le n}\ln p\le n\ln 4$ . It follows that $\frac 12\ln n(\pi(n)-\pi(\sqrt n))\le \sum_{\sqrt n<p<n}\ln p\le n\ln 4$ . But then $\pi(n)\le 2\ln 4\frac{n}{\ln n}+\pi(\sqrt n)\le 2\ln 4\,\frac{n}{\ln n}+\sqrt n< 4\frac{n}{\ln n}$ for large $n$ .

Now let us turn to elementary lower bounds. They are actually not required for the proof of the prime number theorem outlined below, but they may be of some independent interest, especially to those who haven't taken advanced courses in analysis by this moment and may be unable to go through the entire proof yet. We have already seen that

@@ Line 9: / Line 9: @@
 We shall start with some elementary observations due to Chebyshev.
 ==Chebyshev's estimates==
-The key observation is that <math>{2m\choose m}=\frac{(2m)!}{(m!)^2}</math> is divisible by all primes <math>p</math> satisfying <math>m<p\le 2m</math>. Indeed, every such prime appears in the numerator <math>(2m)!</math> and does not appear in the denominator <math>(m!)^2</math>. Thus, <math>\prod_{p\mathrm{\ prime,}\,m<p\le 2m}p\le {2m\choose m}<2^m</math>.
+The key observation is that <math>{2m\choose m}=\frac{(2m)!}{(m!)^2}</math> is divisible by all primes <math>p</math> satisfying <math>m<p\le 2m</math>. Indeed, every such prime appears in the numerator <math>(2m)!</math> and does not appear in the denominator <math>(m!)^2</math>. Thus, <math>\prod_{m<p\le 2m}p\le {2m\choose m}\le 2^{2m}</math> (from now on, <math>p</math> will always stand for a prime number, so, to shorten the notation, we will not explicitly write "<math>p</math> is prime" in the descriptions of sum and product ranges). Similarly, considering <math>{2m+1\choose m}=\frac{(2m+1)!}{m!(m+1)!}</math>, we see that
+<math>\prod_{m+1<p\le 2m+1}p\le {2m+1\choose m}\le 2^{2m}</math> (the last inequality holds because <math>{2m+1\choose m}={2m+1\choose m+1}</math> and <math>{2m+1\choose m}+{2m+1\choose m+1}\le 2^{2m+1}</math>). Now we are ready to prove the
+===Upper bound for the product of primes===
+====Lemma 1====
+For every positive integer <math>n</math>, we have
+<math>\prod_{p\le n}p\le 4^n</math>
+====Proof of Lemma 1====
+[[Induction]] on <math>n</math>. The base <math>n=1</math> is obvious. Suppose now that the statement holds for all numbers strictly less than <math>n</math>. If <math>n=2m</math> is even, then
+<math>\prod_{p\le n}p=\left(\prod_{p\le m}p\right)\cdot \left(\prod_{m<p\le 2m}p\right)\le
+^m 2^{2m}=4^{2m}=4^n.
+</math>
+If <math>n=2m+1</math> is odd, then
+<math>\prod_{p\le n}p=\left(\prod_{p\le m+1}p\right)\cdot \left(\prod_{m+1<p\le 2m+1}p\right)\le
+^{m+1} 2^{2m}=4^{2m+1}=4^n.
+</math>
+----
+From this lemma we can easily derive the
+===Upper bound for the number of primes===
+====Lemma 2====
+<math>\pi(n)\le 4\frac n{\ln n}</math> for large <math>n</math>
+====Proof of Lemma 2====
+Rewrite the inequality of Lemma 1 as <math>\sum_{p\le n}\ln p\le n\ln 4</math>. It follows that
+<math>\frac 12\ln n(\pi(n)-\pi(\sqrt n))\le \sum_{\sqrt n<p<n}\ln p\le n\ln 4</math>.
+But then <math>\pi(n)\le 2\ln 4\frac{n}{\ln n}+\pi(\sqrt n)\le 2\ln 4\,\frac{n}{\ln n}+\sqrt n< 4\frac{n}{\ln n}</math> for large <math>n</math>.
+----
+Now let us turn to elementary lower bounds. They are actually not required for the proof of the prime number theorem outlined below, but they may be of some independent interest, especially to those who haven't taken advanced courses in analysis by this moment and may be unable to go through the entire proof yet. We have already seen that

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