Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm referring to the theorem given here, which is

$$\displaystyle\lim_{n\to \infty} \:\: \left(\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right)\right) \;\;\; = \;\;\; \exp(\hspace{.01 in}\gamma)$$

where $p$ ranges over the primes and $\gamma$ is the Euler-Mascheroni constant.


What I'm really interested in is a good (effective) lower bound for $\;\; \displaystyle\prod_{p\leq n} \: \left(1-\frac1p \right) \;\;$,
so if you have another way to get such a bound, that would work too.


Otherwise, is it explicitly known how large and $n$ is sufficiently large to get $$\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right) \; < \; c$$

where, for example, $\: c=2 \:$ or $\: c = \frac95 \;$?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

See Theorem 6.12 in Pierre Dusart's Estimates of some functions over primes without RH. In particular for all $x \ge 2973$ one has the effective unconditional lower bound: $$\prod_{p\le x}\left(1 - \frac1p\right) > \frac{e^{-\gamma}}{\ln x}\left(1 - \frac{0.2}{\ln^2 x}\right).$$

So, already for $x$ in this range it is true that one can take $c = 1.78666$, which is much smaller than $c=\tfrac95$. A small amount of computation would be all it takes to get the smallest $x$ that works.

Update: it looks like $c=2$ is true for $n\ge 14$, while $c=\tfrac95$ holds for $n \ge 469$.

share|improve this answer

Here is some place to start

$$\prod_{p\leq x}(1-\frac{1}{p})=e^{\sum_{p\leq x}\ln(1-\frac{1}{p})}=e^{\sum_{p\leq x}-\frac{1}{p}-\frac{1}{2p^2}-\frac{1}{3p^3}...}=ce^{-\sum_{p\leq x}\frac{1}{p}}=c'e^{-\ln(\ln(x))}=\frac{c'}{\ln(x)}$$

Where the last estimate comes from $$\sum_{p\leq x} \frac{1}{p}=\ln(\ln(x))+O(1)$$

I think what you should do is bound the terms that aern't reciprocals of primes in the expansion of the logarithm, and then get a good estimate on the sum $\sum_{p\leq x}\frac{1}{p}$ by writing it possibly in terms of a steiljes integral over the prime counting function, and then using a chebyshev estimate for the convergent integral involving the prime counting function, to get an estimate on the constant $c'$.

share|improve this answer
    
Nice answer Ethan. –  mick Nov 11 '13 at 10:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.