I was reading Hanson's proof that $\prod\limits_{p^a \le n}p^a < 3^n$ where $p$ is a prime and it occurred to me that there might be a simpler argument for $\prod\limits_{p \le n} p < 3^n$. Am I wrong?
Here's the argument that I was thinking:
(1) Let $n\# = \prod\limits_{p \le n}p$
(2) For all $n \le 124$, $n\# < 3^n$.
This can be checked manually by comparing $p\#$ to $3^p$ for $p < 124$. In each case $3^p > p\#$
(3) Assume it is true up to some $n-1 \ge 124$
(4) We can assume that $n$ is odd since if it is not $n\# = (n-1)\#$.
(5) There exists $m$ such that $n = 4m-c$ and $c=1$ or $c=3$ with $m \ge 32$
(6) $\dfrac{(4m-c)\#}{m\#} = \dfrac{(4m-c)\#}{(2m)\#}\dfrac{(2m)\#}{(m)\#} < {{4m-c}\choose{2m}}{{2m}\choose{m}}$
Note: For each prime greater than $2m$, it will divide $(4m-c)!$ but not $(2m!)$ or $(2m-c)!$. For each prime $m < p \le 2m$, it will divide $(2m)!$ but not $m!$.
(7) ${{4m-c}\choose{2m}}{{2m}\choose{m}} = \left(\frac{(4m-c)\dots(2m+1)}{(2m-c)!}\right)\left(\frac{(2m)\dots(m+1)}{m!}\right) = {{4m-c}\choose{2m-c,m,m}}$
Where ${{4m-c}\choose{2m-c,m,m}} = \dfrac{(4m-c)!}{(2m-c)!(m!)(m!)}$
(8) Using the Multinomial Theorem:
$${{4m-c}\choose{2m-c,m,m}} = \left(\frac{1}{2^{2m-c}}\right){{4m-c}\choose{2m-c,m,m}}2^{2m-c} < \left(\frac{1}{2^{2m-c}}\right)(1+2)^{4m-c}$$
(9) For both $c=1$ and $c=3$, I can show that for $m\ge 32, \left(\frac{1}{2^{2m-c}}\right)3^{4m-c} < 3^{3m-c}$
- For $c=1$:
$$\frac{3^{4m-1}}{2^{2m-1}} = \left(\frac{27}{2}\right)\left(\frac{81}{4}\right)^{m-1} < 14(21^{m-1})<9(27^{m-1}) = 3^{3m-1}$$
Note: Using induction, it can be show for $m\ge 2, (14)(21^{m-1}) < 9(27^{m-1})$
- For $c=3$:
$$\frac{3^{4m-3}}{2^{2m-3}} = \left(\dfrac{3^9}{2^3}\right)\left(\frac{81}{4}\right)^{m-3} < (2461)(21^{m-3})<(27^{m-3}) = 3^{3m-3}$$
Note: Using induction, it can be show for $m\ge 32, (2461)(21^{m-3}) < 27^{m-3}$
(10) $n\# = (m\#)\left(\frac{(4m-c)\#}{m\#}\right) < (3^m)(3^{3m-c}) = 3^{4m-c} = 3^n$
Edit 1: Great point that I did not make clear that $p$ is only primes.
I have updated my question. To be clear $\prod\limits_{p \le n}p \ne n!$ is referring to the primorial.
Edit 2: Step (7) is completely wrong.
Clearly, if $(3m-c)! > (m!)(2m-c)!$, the value is lower not higher.
Edit 3: I think that I found a fix using the multinomial theorem.
I have corrected step (7) which was previously incorrect.
Please let me know if you see a problem with the updated argument.
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Edit 4: Step(8) is wrong.
To use the mutinomial coefficient, I need three discrete integers not two.
The corrected statement is $(1 + 2 + 1)$ which breaks the argument.