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

Question

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 ''p_n#'' grow according to:

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

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

See also http://math.stackexchange.com/questions/156509/product-of-all-prime-numbers-upto-some-prime-p. — 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 http://math.stackexchange.com/a/174769 ... — draks ..., Nov 25 '13 at 21:31
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. http://www.math.harvard.edu/~elkies/M259.02/chebi.pdf ). — Noam D. Elkies, Nov 25 '13 at 21:47
possible duplicate of Chebyshev: Proof $\prod \limits_{p \leq 2k}{;} p > 2^k$ — Eric Naslund, Nov 26 '13 at 17:34
See this answer http://math.stackexchange.com/questions/33980/how-to-prove-chebyshevs-result-sum-p-leq-n-frac-log-pp-sim-log-n-a/33995#33995 and this answer http://math.stackexchange.com/questions/95766/chebyshev-proof-prod-limits-p-leq-2k-p-2k/95783#95783 — Eric Naslund, Nov 26 '13 at 17:35

score 1 · Accepted Answer · edited Feb 03 '23 at 17:01

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.

edited Feb 03 '23 at 17:01

Glorfindel

3,955

answered Nov 26 '13 at 04:28

Matthew Conroy

11,234

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

1 Answers1