Rate of growth of an Euler Product

Question

Merten's Theorem gives

$$ \prod_{p < x} \left( 1 - \frac{1}{p} \right) \sim k(\log x)^{-1} $$

I also know that

$$ \prod_{p < x} \left(1 - \frac{n}{p} \right) \leq \prod_{p < x} \left(1 - \frac{1}{p} \right)^{n}. $$

My question is, is it nonetheless true that for some constant c we have $$ \prod_{p < x} \left(1 - \frac{n}{p} \right) \sim c(\log(x))^{-n}?$$

Answer 1

There$\let\leq\leqslant\let\geq\geqslant$ is indeed such a relation, with a proof similar to that of Mertens' theorem: $$\prod_{p\leq x}\left(1-\frac1p\right)=e^{-\gamma}(\log x)^{-1}\left(1+O\left(\frac1{\log x}\right)\right).$$

In general, we have $$\prod_{p\leq x}\left(1-\frac rp\right)=c_r(\log x)^{-r}\left(1+O\left(\frac1{\log x}\right)\right)\tag 1$$ for real numbers $r$.

Proof.

We can ignore the factors with $1-\frac rp\leq0$ because these can be absorbed in the constant $c_r$, for sufficiently large $x$. This means we can take logarithms and estimate what's left: $$\sum_{r<p\leq x}\log\left(1-\frac rp\right)=-\sum_{r<p\leq x}\sum_{n\geq1}\frac1n\left(\frac rp\right)^n.$$ Let's concentrate on $\sum_{r<p\leq x}\sum_{n\geq1}\frac1n\left(\frac rp\right)^n$. This is $$\sum_{r<p\leq x}\frac rp+\sum_{r<p\leq x}s_p$$ where $s_p=\sum_{n\geq2}\frac1n\left(\frac rp\right)^n\leq\frac12\sum_{n\geq2}\left(\frac rp\right)^n=\frac{r^2}{2p(p-r)}\ll p^{-2}$, so clearly $\sum_{p>r}s_p$ is absolutely convergent, with sum $S\in\mathbb R$.

The difference between $S$ and the partial sums is of order at most $$\sum_{p>x}s_p=O\left(\sum_{p>x}p^{-2}\right)=O\left(\frac 1x\right).$$

For $\sum_{r<p\leq x}\frac rp$ we have the estimate $r\log\log x+A+O((\log x)^{-1})$ for some $A\in\mathbb R$, by another estimate of Mertens.

Combining our results, we have

$$\begin{align}-\sum_{r<p\leq x}\log\left(1-\frac rp\right)&=r\log\log x+A+O\left(\frac1{\log x}\right)+S+O\left(\frac 1x\right)\\ &=B+r\log\log x+O\left(\frac1{\log x}\right).\end{align}$$

For $y\to0$ we have $|y|\asymp|\log(1+y)|$, which means $$\begin{align}-\sum_{r<p\leq x}\log\left(1-\frac rp\right)&=B+r\log\log x-\log\left(1+O\left(\frac1{\log x}\right)\right).\end{align}$$

Taking $\exp$, $$\prod_{r<p\leq x}\left(1-\frac rp\right)=e^{-B}(\log x)^{-r}\left(1+O\left(\frac1{\log x}\right)\right)$$ which is what we wanted to show, with $c_r=e^{-B}\prod_{p\leq r}\left(1-\frac rp\right)$.

Unfortunately this proof gives us no way to find the constants $c_r$.

Answer 2

$$ \prod_{p < x} \left( 1 - \frac{1}{p} \right) \sim k(\log x)^{-1} \\ \prod_{p < x} \left(1 - \frac{n}{p} \right) \leq \prod_{p < x} \left(1 - \frac{1}{p} \right)^{n}. \\ \implies \prod_{p < x} \left(1 - \frac{n}{p} \right) \leq \prod_{p < x} \left(1 - \frac{1}{p} \right)^{n}\sim k^n(\log x)^{-n}\\ \implies \text{If $c:=k^n$, }\prod_{p < x} \left(1 - \frac{n}{p} \right) \leq c(\log x)^{-n}$$