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I'm writing my bachelor thesis about Brun's sieve method and his theorem.

In one proof I found this statement without further explanation. It is important to show that the product doesn't converge "too quickly" to zero, to show the rest of the theorem.

The full statement is that, for $y \rightarrow \infty$, it holds that $$\prod_{2 < p \leq y}\left(1-\frac{2}{p}\right)\sim\frac{D}{\log ^2 y}.$$ Needless to say that $p$ is meant to be prime here. The $D$ is not mentioned at any other point, so i assume it's some kind of constant.

I tried to find an Euler Product claiming the same, without success. I even digged out some of my complex analysis scripts about the convergence of products, but thats not even what I need to show here.

I would be very happy about every hint or approach anyone can give me here.

EDIT: One Euler Product says that $$\prod_p (1-p^{-s}) = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} = \zeta (s)^{-1}.$$ Does that help in any way? For me it only tells me the convergence to zero.

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marked as duplicate by Eric Naslund, Daniel Fischer Apr 26 '16 at 15:34

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    $\begingroup$ Because $$\prod_{2 < p \leqslant y} \left(1-\frac{2}{p}\right)\approx\exp\left(-2\sum_{2 < p \leqslant y}\frac{1}{p}\right)$$ and one can estimate very precisely the sum in the exponential. $\endgroup$ – Did Apr 24 '16 at 10:57
  • $\begingroup$ @Did : $\prod_{p < y} (1-2/p) = \exp\left(\mathcal{O}(C+y^{-1/2+\epsilon}- 2 \sum_{p < y} 1/p\right) $ where $C = -\sum_{p^k, k \ge 2} \frac{1}{k p^{k}}$ $\endgroup$ – reuns Apr 24 '16 at 11:05
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    $\begingroup$ and yes of course the prime number theorem relies on $\zeta(s)$ and the Euler product, and estimating $\sum_{p < y} 1/p \sim \ln \ln(y)$ is precisely the prime number theorem $\endgroup$ – reuns Apr 24 '16 at 11:07
  • $\begingroup$ So complex analysis might really have helped me :) So do I get it right, that Meissel-Mertens tells me, that $\sum_{p<n} \frac{1}{p} - \log \log n$ converges and therefore there exists a constant $D'$ so that $\frac{D' \cdot \sum_{p<n} \frac{1}{p}}{\log \log n} \rightarrow 1$? $\endgroup$ – RoyPJ Apr 24 '16 at 11:12
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    $\begingroup$ once you proved that it converges, it is clear that $\frac{\sum_{p < y} 1/p}{\ln \ln y} \to 1$, but the PNT just says that $\frac{\sum_{p < y} 1/p}{\ln \ln y} \to 1$, equivalently that $\ln \zeta(s) + \ln(s-1)$ is holomorphic for $Re(s) \ge 1$ $\endgroup$ – reuns Apr 24 '16 at 11:14
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First, taking logarithm of the product and using estimates from Mertens' 2nd theorem, we get \begin{align*} \log[\prod_{2<p\le y}\left(1-\frac{2}{p}\right)] &=\sum_{2<p\le y}\log(1-\frac{2}{p})\\ &=-\sum_{2<p\le y}\frac{2}{p}-\sum_{2<p\le y}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j\\ &=-2\left(\log\log y+M-\frac{1}{2}+\mathcal{O}(\frac{1}{\log y})\right) -\sum_{p>2}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j +\sum_{p>y}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j, \end{align*} where $M$ is the Meissel–Mertens constant.

For the part of double summation, note that \begin{align*} |\sum_{p>y}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j| &<\sum_{p>y}\sum_{j\ge 2}(\frac{2}{p})^{j}\\ &=\sum_{p>y}\frac{4p^{-2}}{1-2p^{-1}}\\ &=\mathcal{O}\left(\sum_{p>y}\frac{1}{p^2}\right)\\ &=\mathcal{O}\left(\sum_{n>y}\frac{1}{n^2}\right)\\ &=\mathcal{O}(\frac{1}{y}), \end{align*} so the sum $$ \sum_{p>2}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j $$ converges.

Let $A$ denote the constant terms $$ A=1-2M-\sum_{p>2}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j. $$

We have \begin{align*} \log[\prod_{2<p\le y}\left(1-\frac{2}{p}\right)] &=-2\log\log y+A+\mathcal{O}(\frac{1}{\log y}). \end{align*}

Taking the exponentials of both sides, we get \begin{align*} \prod_{2<p\le y}\left(1-\frac{2}{p}\right) &=\exp\left(-2\log\log y+A+\mathcal{O}(\frac{1}{\log y})\right)\\ &=\frac{e^A}{(\log y)^2}\left(1+\mathcal{O}(\frac{1}{\log y})\right). \end{align*} Thus we've proved your conclusion with $D=e^A.$

For the constant $A$, we can rewrite it in a much nicer form as suggested by Eric Naslund, \begin{align*} A&=1-2M-\sum_{p>2}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j\\ &=1-2\left[\gamma+\sum_{p}[\log(1-\frac{1}{p})+\frac{1}{p}]\right]+\sum_{p>2}\log(1-\frac{2}{p})+\frac{2}{p}\\ &=1-2\gamma-2(\log\frac{1}{2}+\frac{1}{2})+\sum_{p>2}[\log(1-\frac{2}{p})-2\log(1-\frac{1}{p})]\\ &=-2\gamma+\log 4+\log\Pi_2, \end{align*} so that we have $D=4\Pi_2 e^{-2\gamma}.$

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  • $\begingroup$ Wow, thank you very much for that detailed and quite understandable answer! I can proudly state that I understood every detail of it, what gives me a good feeling on continueing my thesis. $\endgroup$ – RoyPJ Apr 24 '16 at 15:28
  • $\begingroup$ The most part of this could be replaced by the double inequality $$e^{-4/p^2}\leqslant\left(1-\frac2p\right)e^{2/p}\leqslant1,$$ losing only the effective value of the constant $D$ in the assertion to be proved, but not the existence of $D$. $\endgroup$ – Did Apr 24 '16 at 15:31
  • $\begingroup$ @RoyPJ, you're welcome. Hope you can finish your thesis with distinction. $\endgroup$ – Zhenhua Liu Apr 24 '16 at 23:04
  • $\begingroup$ The exact constant $D=e^A$ is given by $$D=4\Pi_2 e^{-2\gamma}$$ where $\Pi_2$ is the twin prime constant and $\gamma$ is the Euler-Mascheroni constant. This is proven in the accepted answer here: math.stackexchange.com/questions/22411/… $\endgroup$ – Eric Naslund Apr 26 '16 at 12:03
  • $\begingroup$ @Eric Naslund, thank you for the suggestion. $\endgroup$ – Zhenhua Liu Apr 26 '16 at 12:37

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