5
$\begingroup$

Does anyone know how I could get a good upper bound for the following:

$$R := \prod_{\substack{ p \; \text{prime} \\ 5 \leq p < n}}p^{\frac{n}{p-1}}$$

I'm not that skilled at asymptotic analysis and the best I have been able to come up with is $\exp\left({\frac{1.01624n^2}{4}}\right)$ since $p-1 \geq 4$ for all $p \geq 5$ and:

$$\log R = n \sum_{\substack{ p \; \text{prime} \\ 5 \leq p <n}}{\frac{1}{p-1}} \log p < \frac{n}{4} \sum_{\substack{ p \; \text{prime} \\ 5 \leq p <n}} \log p < \frac{1.01624n^2}{4}$$

since the first Chebyshev function is bounded above by $1.01624n$ (I saw this bound on Wikipedia). I know that this bound is way too large though since it does not take into account that $\lim_{p \to \infty} \frac{1}{p-1} = 0$. Does anyone have any other suggestions? Thanks for your help.

$\endgroup$

1 Answer 1

2
$\begingroup$

Mertens second formula states $$\sum_{p\leq x}\frac{\log\left(p\right)}{p}=\log\left(x\right)+O\left(1\right)$$ so $$\log\left(\prod_{5\leq p\leq n-1}p^{\frac{n}{p-1}}\right)=n\sum_{5\leq p\leq n-1}\frac{\log\left(p\right)}{p-1}=n\sum_{p\leq n}\frac{\log\left(p\right)}{p}+O\left(n\right)=n\log\left(n\right)+O\left(n\right)$$ hence $$\prod_{5\leq p\leq n-1}p^{\frac{n}{p-1}}=e^{n\log\left(n\right)+O\left(n\right)}.$$ Expanding a little bit, the sositution holds because$$\sum_{5\leq p\leq n-1}\frac{\log\left(p\right)}{p-1}-\sum_{p\leq n-1}\frac{\log\left(p\right)}{p}=\sum_{p\leq n-1}\frac{\log\left(p\right)}{p\left(p-1\right)}-\frac{\log\left(3\right)}{2}\leq\sum_{p\geq2}\frac{\log\left(p\right)}{p\left(p-1\right)}-\frac{\log\left(3\right)}{2}=O\left(1\right)$$ because $\sum_{p\geq2}\frac{\log\left(p\right)}{p\left(p-1\right)}$ is convergent. So$$n\sum_{5\leq p\leq n-1}\frac{\log\left(p\right)}{p-1}=n\sum_{p\leq n-1}\frac{\log\left(p\right)}{p}+O\left(n\right)=n\sum_{p\leq n}\frac{\log\left(p\right)}{p}+O\left(n\right)$$ because adding the term (it's not necessary, you can find the asymptotic with $n-1$) $\log\left(n\right)/n$ gives an error of $$n\frac{\log\left(n\right)}{n}=\log\left(n\right)=O\left(n\right).$$

$\endgroup$
2
  • $\begingroup$ Thank you. This is probably a stupid question, but how can you change the denominator in the summation from $p-1$ to $p$ with the addition of a constant term? Couldn't it work our to be more than a constant (especially considering the factor of $n$ outside the summation)? Thanks. $\endgroup$
    – Ari
    Mar 5, 2015 at 23:57
  • $\begingroup$ @Ari There was an error, thank you. I added more details, hope it's more clear now. $\endgroup$ Mar 6, 2015 at 8:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .