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.


1 Answer 1


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 $$ \vartheta(x) =\sum_{p\le x} \log p.$$

This yields your required bounds.


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