Estimates for a product involving primes

Question

Is $$\prod_{p\leq z}\bigg(1 + \frac{p}{(p-1)}\frac{\log z}{\log p}\bigg) = O(z)?$$ where $p$ ranges over the primes less than $z$. I believe I can show that it is $O(z\log \log z)$.

Answer 1

Let $f(z)=\exp\left((\log z)^{1/2}\right),$ and consider$$\prod_{p\leq f(z)}\left(1+\frac{\log z}{\log p}\right).$$ This is clearly a lower bound for your product, and this is bounded below by $$\prod_{p\leq f(z)}\left(\frac{\log z}{\log f(z)}\right)=\left(\log z\right)^{\pi(f(z))/2}=\exp\left(\frac{1}{2}\pi(f(z))\log\log z\right).$$ Now, $$\pi(f(z))=\pi\left(e^{\sqrt{\log z}}\right)\gg_{A}(\log z)^{A}$$ for any $A$ , and hence the product is $$\geq\exp\left(C_{A}(\log z)^{A}\right)$$ for any $A>0$ where $C_{A}>0$ is a constant depending on $A$ . Hence the product cannot be $O(z)$ or even $O(z\log\log z)$ as the above quantity grows faster than $z^{k}$ for any $k>0$ .

I believe that if you are careful, then you can show that this function grows like $$e^{Cz(1+o(1)}.$$