Bertrand's Conjecture

Theorem (Bertrand's Conjecture)
Fir  
\[n \ge 1\]
  there is at least one prime number between  
\[n\]
  and  
\[2n\]
  inclusive.
Proof
Proof. Suppose to the contrary that there is some  
\[n\]
  with no prime between  
\[n\]
  and  
\[2n\]
. Consider the prime factors of  
\[C_n = \begin{pmatrix}2n\\n\end{pmatrix}= \frac{(2n)!}{n!n!}\]
. If  
\[p\]
  is such a factor with  
\[\frac{2n}{3} \lt p \lt n\]
  then  
\[\left[ \frac{2n}{p} \right] =2 \left[ \frac{n}{p} \right] =2\]
, so  
\[p\]
  is not a factor of  
\[2n\]
, so  
\[p \le \frac{2n}{3}\]
/
We need the following lemma.
Lemma
For any integer  
\[\]
, none of the prime powers in the prime factorisation of  
\[2n\]
.
For example, if  
\[n=4\]
,  
\[C_n=70=2^1 \times 5^1 \times 7^1\]
. None of  
\[2^1, \; 5^1, \; 7^1\]
  exceeds  
\[2n=8\]
/
Proof
Let  
\[v_p(n)\]
  be the number of times a prime  
\[p\]
  occurs in the prime factorisation of  
\[n!\]
  is  
\[\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] +...\]
.
\[v_p(C_n)=v_p((2n)!)-2 v_p(n!)= \left( \left[ \frac{2n}{p} -2 \frac{n}{p} \right] \right) + \left( \left[ \frac{2n}{p^2} -2 \frac{n}{p^2} \right] \right) + \left( \left[ \frac{2n}{p^3} -2 \frac{n}{p^3} \right] \right) +...\]

\[p^k \gt 2n \rightarrow \left( \left[ \frac{2n}{p^k} -2 \frac{n}{p^k} \right] \right)=0\]
  else this term is at most 1. Hence  
\[v_p(C_n)\]
  is at less than or equal to the largest  
\[k\]
  satisfying  
\[p^k \le 2n \rightarrow p^{v_p(C_n)} \le 2n\]
.
Let  
\[n \gt 4\]
  (
\[n=1, \; 2, \; 3, \; 4\]
  are not counterexamples), then  
\[\sqrt{2n} \lt \frac{2n}{3}\]
. We can divide factors of  
\[C_n\]
  into those less than  
\[\sqrt{2n}\]
  and those between  
\[\sqrt{2n}\]
  and  
\[\frac{2n}{3}\]
.
\[C_n=\underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \underbrace{p_{r+1}^{a_{r+1}}...p_k^{a_k}}_{\sqrt{2n} \lt p_i \lt \frac{2n}{3}}\]
.
By the above lemma none of the primes in the first factor on the right hand side exceeds  
\[2n\]
  so  
\[\underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \le (2n)^{\sqrt{2n}}\]
.
The primes in the second factor all have power equal to 1, since  
\[p \gt \sqrt{2n} \rightarrow p^2 \gt 2n\]
  and the power could not be 2 or greater again by the above lemma. It then follows that  
\[\underbrace{p_{r+1}^{a_{r+1}}...p_k^{a_k}}_{ \sqrt{2n} \lt p_i \lt \frac{2n}{3}} \le (\frac{2n}{3})! \le 4^{2n/3}\]
.
We need another lemma.
Lemma 2
\[\frac{4^n}{2n} \le C_n\]

Proof
\[4^n \le (1+1)^{2n} = \sum_{k=0}^{2n} \begin{pmatrix}2n\\k\end{pmatrix}=2+ \sum_{k=1}^{2n-1} \begin{pmatrix}2n\\k\end{pmatrix} \le 2+(2n-1) \begin{pmatrix}2n\\n\end{pmatrix} \le 2n \begin{pmatrix}2n\\n\end{pmatrix}\]

Hence  
\[\frac{4^n}{2n} \le C_n\]

With this lemma we have  
\[\frac{4^n}{2n} \le C_n= \underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \underbrace{p_{r+1}^{a_{r+1}}...p_K^{r_k}}_{ \sqrt{2n} \lt p_i \lt \frac{2n}{3}} \le (2n)^{\sqrt{2n}} 4^{2n/3}\]
.
This can be shown to be true for  
\[n=1, \; 2,...,467\]
  but false for  
\[n \ge 468\]
.To show that there are no coun terexamples for  
\[n le 467\]
 , we find a sequence of primes beginning  
\[2 \le p \le 467\]
  such that each prime is no more than twice the previous one. Here is such a sequence:
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631