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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 $\: c = \frac95 \;$?

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up vote 6 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 $n \ge 469$.

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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 $c'$.

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Nice answer Ethan. – mick Nov 11 '13 at 10:54

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