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