Asymptotic behaviour of $\prod_{p \leq x} (1 + 4/(3p) + C p^{-3/2})$

Question

I'm reading Montgomery and Vaughan and in it they state quite simply \begin{equation} \prod_{p \leq x} \left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}} \right) \ll (\log x)^{4/3} \end{equation} as $x \rightarrow \infty$ and where $C$ is some constant. It states that it's using Mertens' formula so there should probably be some relation of the form \begin{equation} \prod_{p \leq x} \frac{1 + \frac{4}{3p} + \frac{C}{p^{3/2}}}{\left(1 - \frac{1}{p}\right)^{4/3}} \ll 1 \end{equation}

but I can't quite get it.

Answer 1

Actually the relation you wanted to write down, via Merten's $3$rd theorem, should be

$$\prod_{p\le x}\left(1-\frac{1}{p}\right)^{4/3}\left(1+\frac{4}{3p}+\frac{C}{p^{3/2}}\right) \ll 1.$$

Hint: we know $(1+x)^\alpha=1+\alpha x+O(x^2)$ (for $x\approx0$) and when $\prod(1-p^{-s})$ converges.

Answer 2

Basically you should just mimic the proof of Mertens' formula. We have that \[\sum_{p \leq x} \log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) = \frac{4}{3} \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x}\left(\log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) - \frac{4}{3p}\right),\] and the first term is asymptotic to $\frac{4}{3} \log \log x$, while one can write the logarithm in terms of its power series in order to show that the second term is $o(1)$, from which the result follows.