Editing Proof of Bertrand's postulate (section)

==Proof of Bertrand's Postulate==
Assume that there is a [[counterexample]]: an integer ''n''&nbsp;≥&nbsp;2 such that there is no prime ''p'' with ''n''&nbsp;<&nbsp;''p''&nbsp;<&nbsp;2''n''.

If 2 ≤ ''n'' &lt; 630, then ''p'' can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being the largest prime less than twice its predecessor) such that ''n''&nbsp;<&nbsp;''p''&nbsp;<&nbsp;2''n''. Therefore, ''n''&nbsp;≥&nbsp;630.

There are no prime factors ''p'' of <math>\textstyle\binom{2n}{n}</math> such that:
* 2''n'' < ''p'', because every factor must divide (2''n'')!;
* ''p'' = 2''n'', because 2''n'' is not prime;
* ''n'' < ''p'' < 2''n'', because we assumed there is no such prime number;
* 2''n''/3 < ''p'' ≤ ''n'': by [[#Lemma 3|Lemma 3]].
Therefore, every prime factor ''p'' satisfies ''p'' ≤ 2''n''/3.

When <math>p > \sqrt{2n},</math> the number <math>\textstyle {2n \choose n} </math> has at most one factor of ''p''. By [[#Lemma 2|Lemma 2]], for any prime ''p'' we have ''p''<sup>''R''(''p'',''n'')</sup> ≤ 2''n'', and <math>\pi(x)\le x-1</math> since 1 is neither prime nor composite. Then, starting with [[#Lemma 1|Lemma 1]] and decomposing the {{nowrap|right-hand}} side into its prime factorization, and finally using [[#Lemma 4|Lemma 4]], these bounds give:
:<math>\frac{4^n}{2n}\le\binom{2n}{n}=\left(\,\prod_{p\,\le\,\sqrt{2n}}p^{R(p,n)}\right)\!\!\left(\prod_{\sqrt{2n}\,<\,p\,\le\,2n/3}\!\!\!\!\!\!\!p^{R(p,n)}\right)<\left(\,\prod_{p\,\le\,\sqrt{2n}}\!\!2n\right)\!\!\left(\prod_{p\,\le\,2n/3}\!\!p\right)\le(2n)^{\sqrt{2n}-1}4^{2n/3}.</math>
Therefore
:<math>4^{n/3}<(2n)^\sqrt{2n}</math>, which simplifies to <math>2^\sqrt{2n}<(2n)^3.</math>
Taking the base-2 [[logarithm]] of both sides yields
:<math>\sqrt{2n}<3\log_2(2n).</math>
By [[concave function|concavity]] of the right-hand side as a function of ''n'', the last inequality is necessarily verified on an interval. Since it holds true for <math>n=426</math> and it does not for <math>n=427</math>, we obtain
:<math>n<427.</math>
But these cases have already been settled, and we conclude that no counterexample to the postulate is possible.

===Addendum to proof===
It is possible to reduce the bound to <math>n=50</math>.

For <math>n\ge17,</math> we get <math>\pi(n)<\frac{n}{2}-1</math>, so we can say that the product <math>p^R</math> is at most <math>(2n)^{0.5\sqrt{2n}-1}</math>, which gives

:<math>\begin{align}&\frac{4^n}{2n}\le\binom{2n}{n}\le(2n)^{0.5\sqrt{2n}-1}4^{2n/3}\\&4^{\sqrt{2n}}\le(2n)^3\\&2\sqrt{2n}\le3\log_2(2n)\end{align}</math>

which is true for <math>n=49</math> and false for <math>n=50</math>.