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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.

Did
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RoyPJ
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    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. – Did Apr 24 '16 at 10:57
  • @Did : $\prod_{p < y} (1-2/p) = \exp\left(\mathcal{O}(C+y^{-1/2+\epsilon}- 2 \sum_{p < y} 1/p\right) $ where $C = -\sum_{p^k, k \ge 2} \frac{1}{k p^{k}}$ – reuns Apr 24 '16 at 11:05
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    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 – reuns Apr 24 '16 at 11:07
  • 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 $\frac{D' \cdot \sum_{p<n} \frac{1}{p}}{\log \log n} \rightarrow 1$? – RoyPJ Apr 24 '16 at 11:12
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    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 $Re(s) \ge 1$ – reuns Apr 24 '16 at 11:14
  • you should try reading on "zeta riemann zero-free region" for example http://math.stackexchange.com/questions/1079485/importance-of-the-zero-free-region-of-riemann-zeta-function and see how the Laplace/Fourier/Mellin transform helps – reuns Apr 24 '16 at 11:18
  • First of all thank you for your very helpful comments. Indeed i even remember those properties of the zeta function from one lecture and i could use it to prove it. But given that i haven't used the zeta function in my thesis yet i would rather use the prime number theorem. Please don't be mad at me for having a mindblock on how $\pi(x)$ ~ $ \frac{x}{\log x}$ is equivalent to the here stated convergence.. – RoyPJ Apr 24 '16 at 11:30
  • @user1952009 Sure you first comment is addressed at me? – Did Apr 24 '16 at 12:26
  • $\frac{1}{s}\ln(\zeta(s))$ is the Mellin transform of $\pi(x)$ plus a function holomorphic for $Re(s) > 1/2$, and $\frac{1}{s+1}\ln(\zeta(s+1))$ is the Mellin transform of $\sum_{p \le x} 1/p$ plus a function holomorphic for $Re(s) > -1/2$, – reuns Apr 24 '16 at 13:06
  • I mean $\frac{1}{s} \ln(s+1)$ is the Mellin transform of $\sum_ {p \le x} 1/p$ plus an holomorphic function. now $\ln(s-1)$ is the Mellin transform of $\frac{x}{\ln x}1_{x \ge 1}$ hence $\ln s$ is the Mellin transform of $\frac{1}{\ln x} 1_{x \ge 1}$, and $\frac{1}{s}\ln(s)$ is the Mellin transform of $\int_1^x \frac{dy}{\ln y}$ which should be $\sim \ln \ln x$, hence $\pi(x) \sim \frac{x}{\ln x}$ implies $\sum_{p \le x} 1/p \sim \ln \ln x$ (if I didn't do anything wrong) – reuns Apr 24 '16 at 13:12
  • @user1952009 Once again, whom are you talking to? And how are your three last comments actually addressing the question at hand? – Did Apr 24 '16 at 14:48
  • @user1952009 Sorry but none of the mass of (true and certainly interesting by themselves) results you are pouring into this thread is necessary to solve the question at hand, just like the estimate in your very first comment is not necessary to explain the assertion in my comment, $\approx$ included. – Did Apr 24 '16 at 14:56
  • @user1952009 The question asked is not to prove Mertens 2 (and if it was, your comments would not allow to do that). Re the answer below, its most lengthy part could be replaced by the simple inequalities $$-x-x^2\leqslant\log(1-x)\leqslant-x,$$ valid for every $0<x<2/3$. (If ever you answer to this comment (which is probably not necessary), I suggest to omit the rhetorical "I don't understand what you mean", having used it twice is already more than enough.) – Did Apr 24 '16 at 15:08
  • @user1952009 Please stick to the maths. Comment flagged. (You now added to your comment a mathematical assertion... unfortunately it is offtopic.) – Did Apr 24 '16 at 15:11

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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 $D=4\Pi_2 e^{-2\gamma}.$

  • 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. – RoyPJ Apr 24 '16 at 15:28
  • 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 $D$. – Did Apr 24 '16 at 15:31
  • @RoyPJ, you're welcome. Hope you can finish your thesis with distinction. –  Apr 24 '16 at 23:04
  • 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: http://math.stackexchange.com/questions/22411/computing-the-product-of-p-p-2-over-the-odd-primes/22435#22435 – Eric Naslund Apr 26 '16 at 12:03
  • @Eric Naslund, thank you for the suggestion. –  Apr 26 '16 at 12:37
  • That's interesting. Does anybody know if this is what Brun found out in his thesis from 1919 or if this was found later on? I'd like to put it in the historical context. Thank you all very much. – RoyPJ Apr 30 '16 at 08:01
  • Nevermind, ofc this is the definition of the twin prime constant... – RoyPJ Apr 30 '16 at 08:04