# Prove that $2^x < \prod_{p\le x} p < (13/4)^x$ for sufficiently large x

Prove that $2^x < \prod_{p\le x} p < (13/4)^x$ for sufficiently large x. Here $p$ is prime.

So if we take logs we need to show for sufficiently large x, $x\log 2 < \sum_{p\le x}\log p < x\log(13/4)$. Also according to http://en.wikipedia.org/wiki/Primorial

Asymptotically, primorials ''pn#'' grow according to:

$p_n\# = e^{(1 + o(1)) n \log n}$

where $o(\cdot)$ is the little o notation.

• – Dietrich Burde Nov 25 '13 at 21:20
• ...or you may use $\sum_{k=1}^n \log p_n = \int_2^n \log k\; d\pi(k)$ and put in your favorite representation for $\pi(x)$. See here math.stackexchange.com/a/174769 ... – draks ... Nov 25 '13 at 21:31
• why is this contest-math? – draks ... Nov 25 '13 at 21:35
• Because the difficulty level of this problem is at least contest-math level, yet it's easy to understand. – Christmas Bunny Nov 25 '13 at 21:45
• I guess that it's "contest-math" because it came from a math contest and thus calls for an elementary proof along the lines of Čebyšev's method (see e.g. math.harvard.edu/~elkies/M259.02/chebi.pdf ). – Noam D. Elkies Nov 25 '13 at 21:47

Rosser and Schoenfeld prove, in their Theorem 4, that $$x\left(1-\frac{1}{2 \log x}\right) < \vartheta(x) < x \left( 1+\frac{1}{2 \log x} \right)$$ for $x\ge 563$, where $$\vartheta(x) =\sum_{p\le x} \log p.$$