<s>[INST] Mertens' theorem with the numerator other than $1$ The Mertens' theorem is of the form
$$\prod_{p\le X}\left(1-\frac{1}{p}\right)=\frac{e^{-\gamma}}{\log X}\left\{1+o\left(\frac{1}{\log X}\right)\right\}.$$
I want to replace $1$ with others, for example $2$. At the first glance I think that (and I believe I have seen it somewhere)
$$\prod_{2\neq p\le X}\left(1-\frac{2}{p}\right)\sim\prod_{p\le X}\left(1-\frac{1}{p}\right)^2.$$
However, the following arguments (that I tried) give
$$\prod_{2\neq p\le X}\left(1-\dfrac{2}{p}\right)=\left(\dfrac{2e^{-\gamma}}{\log X}\right)^2\prod_{p>2}\left(1-\dfrac{1}{(p-1)^2}\right)\left\{1+o\left(\dfrac{1}{\log X}\right)\right\},$$
This suggest another constant in the middle above that I'm not sure I did it right. Here are the arguments I mentioned.
We first have for $p\neq 2$,
$$\log\left(1-\frac{2}{p}\right)=\log\left(1-\frac{1}{p}\right)^2-\log\left(1+\frac{1}{p(p-2)}\right).$$
Thus, by Mertens' theorem (with P.N.T. form) that
$$\sum_{p\le X}\log\left(1-\frac{1}{p}\right)^{-1}=\log\log X+\gamma+o\left(\frac{1}{\log X}\right),$$
\begin{align*}
\sum_{2< p\le X}\log\left(1-\frac{2}{p}\right) &= \sum_{2< p\le X}2\log\left(1-\frac{1}{p}\right)-\sum_{2< p\le X}\log\left(1+\frac{1}{p(p-2)}\right)\\
&= -2\log\log X-2\gamma+\log 4+o\left(\frac{1}{\log X}\right)+\sum_{2< p\le X}\log\left(1-\frac{1}{(p-1)^2}\right).
\end{align*}
Also, $$\sum_{2< p\le X}\log\left(1-\frac{1}{(p-1)^2}\right)=c+O\left(\sum_{p>X}\log\left(1-\frac{1}{(p-1)^2}\right)\right)=c+O(1/X),$$
where $c=\sum_{p>2}\log\Big(1-1/(p-1)^2\Big)$.
Therefore, we obtain
$$\prod_{2<p\le X}\left(1-\frac{2}{p}\right)=\left(\frac{2e^{-\gamma}}{\log X}\right)^2\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)\left\{1+o\left(\frac{1}{\log X}\right)\right\},$$
as desired.
 [/INST] That first asymptotic relation is incorrect.
For $p \geq 3$,
$$
\frac{(1-1/p)^2}{1-2/p} = \frac{1-2/p+ 1/p^2}{1-2/p} = 1 + \frac{1}{p(p-2)}
$$
and $\prod_{p\geq 3} (1 + \frac{1}{p(p-2)})$ converges by an argument similar to convergence of $\prod_{p} (1 + \frac{1}{p^2})$.  So
$$
\prod_{3 \leq p \leq X} \left(1 - \frac{1}{p}\right)^2 \sim 
c\prod_{3 \leq p \leq X}\left(1 - \frac{2}{p}\right) 
$$
for the constant $c = \prod_{p \geq 3} (1 + \frac{1}{p(p-2)}) > 1$.
</s>