Is there a bound for $$\prod_{i=1}^{m}\Big(1-\frac{1}{p_i}\Big)$$ where $p_i$ is $i$th prime?
What if $m=O(\log n)$?
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do you have Hardy and Wright? – Will Jagy May 29 '16 at 21:08
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@WillJagy actually no. what is this product called? – May 29 '16 at 21:13
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4Mertens' third theorem – Daniel Fischer May 29 '16 at 21:28
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Since $\sum_{p}\frac{1}{p}$ is divergent (see Erdos' elementary proof), your product is convergent to zero as $m\to +\infty$, and since $\sum_{p\leq x}\frac{1}{p}\approx \log\log x$, the product behaves like $\frac{1}{\log m}$. – Jack D'Aurizio May 29 '16 at 21:59
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Let $n_i$ be such that $p_i^{n_i}\ge p_{m+1}$ For all $n$ we have $\frac1{1-\frac1{p_i}}\ge 1+p_i^{-1}+\ldots+p_i^{-n}$. As every natural number, less than $p_{m+1}$ can be uniquely written as product of powers of $p_i$s, $$\prod_{i=1}^m \left(\frac1{1-\frac1{p_i}}\right)\ge \sum_{i=1}^{p_{m+1}-1} \frac1i\ge 1+\log p_{m+1}$$ So, your product is can be bounded by $$\prod_{i=1}^m \left(1-\frac1{p_i}\right)\le \frac1{1+\log p_{m+1}}$$ Using PNT, $\log p_{m+1}\sim \log m-\log\log m$. So, $m\sim \log n$ gives $\log p_{m+1}\sim \log\log n-\log\log\log n$
Emre
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I feel the same. But, I expect any other bound to be around $1/\log {p_m}$. From Daniel's comment, the bound is $e^{-\gamma}/\log {p_m}$. – Emre May 29 '16 at 21:28
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@E.Girgin : $\log (\prod_{p \le x} 1-1/p) = \sum_{p \le x} \sum_{k=1}^\infty (-1) \frac{p^k}{k} \approx - \sum_{p \le x} \frac{1}{p}- \lim_{s \to 1} \ln (\zeta (s-1) - \frac{1}{s-1}) $ $= - \sum_{p \le x} \frac{1}{p} - \gamma \approx -\ln\ln(x)- \gamma$ – reuns May 29 '16 at 21:45
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Hint
It is known that
$$\dfrac{\phi(n)}n = \prod\limits_{p\,|\, n}\left(1-\dfrac1p\right).$$
Required production equals to $\dfrac{\phi(p_n\#)}{p_n\#}.$
And I can recommend light and detail info.
Yuri Negometyanov
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