Estimate the product!

We are going to prove that $B$ is always greater than $A$ whenever $n > 1$ .

For the proof we are using induction on $n$ . The base case $n=2$ is clear, so assume we have already proven the statement for every $2 \leq m < n$ integer.

If $n$ is an even number, it is certain that $\displaystyle \prod_{\substack{p \leq n-1 \\ p \ prime}} p = \displaystyle \prod_{\substack{p \leq n \\ p \ prime}} p$ , so according to the induction step $\displaystyle \prod_{\substack{p \leq n \\ p \ prime}} p = \displaystyle \prod_{\substack{p \leq n-1 \\ p \ prime}} p < 4^{n-1} < 4^n$ which verifies the induction statement for $n$ . Therefore we can suppose that $n$ is odd, hence it is in the form of $2k+1$ for $k$ being a positive integer.

Firstly, notice that our product can be split in the following way:

$\displaystyle \prod_{\substack{p \leq 2k+1 \\ p \ prime}} p = \displaystyle \prod_{\substack{p \leq k+1 \\ p \ prime}} p \ \cdot \displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p$

Next recall the well known fact that if $a | b$ , $\gcd(a,c)=1$ and $\frac{b}{c}$ is an integer, we have $a | \frac{b}{c}$ . From this we obtain

$\displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p \ | \ \frac{(2k+1) \cdot 2k \cdot (2k-1) \cdot \ldots \cdot (k+2)}{k \cdot (k-1) \cdot \ldots \cdot 2 \cdot 1} = {2k+1 \choose k}$

since $\displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p | (2k+1) \cdot 2k \cdot (2k-1) \cdot \ldots \cdot (k+2)$ , $\gcd(\displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p \ , \ k \cdot (k-1) \cdot \ldots \cdot 2 \cdot 1)=1$ and ${2k+1 \choose k}$ is clearly an integer.

Hence it follows that $\displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p \leq {2k+1 \choose k}$ .

Furthermore consider the following binomial expansion:

$2 \cdot 4^k = 2^{2k+1} = (1+1)^{2k+1} = {2k+1 \choose 0} + {2k+1 \choose 1} + \ldots + {2k+1 \choose k} + {2k+1 \choose k+1} + \ldots + {2k+1 \choose 2k+1} > {2k+1 \choose k} + {2k+1 \choose k+1} = 2 \cdot {2k+1 \choose k}$

hence $4^k > {2k+1 \choose k}$ , which holds for any $k > 0$ integer.

From this and the induction step we can conclude that

$\displaystyle \prod_{\substack{p \leq n \\ p \ prime}} p = \displaystyle \prod_{\substack{p \leq 2k+1 \\ p \ prime}} p = \displaystyle \prod_{\substack{p \leq k+1 \\ p \ prime}} p \ \cdot \displaystyle \prod_{\substack{k+2 \leq p \leq 2k+1 \\ p \ prime}} p < 4^{k+1} \cdot {2k+1 \choose k} < 4^{k+1} \cdot 4^k = 4^{2k+1} = 4^n$ .

Henceforth our induction is complete.

Since $\ln A(n) \; = \; \sum_{p \le n} \ln p \; \le \; \psi(n)$ where $\psi$ is the second Chebyshev function, and the Prime Number Theorem is equivalent to the statement that $n^{-1}\psi(n) \to 1$ as $n \to \infty$ , we deduce that $\lim\,\sup\,n^{-1}\ln A(n) \,\le \, 1 \, < \, \ln4$ , and so $\lim_{n \to \infty} \frac{A(n)}{B(n)} = 0$ .