As$\ n \to \infty$, how does a product over the primes less than$\ p_n$ equal the same product over the primes less than$\ n$?

Question

How is$$\ \lim_{x\to \infty} \log \log x \prod_{i< \log x} \frac{p_i -1}{p_i}= \\ \lim_{x\to \infty} \log \log x \prod_{p < \log x}_{p prime} \frac{p-1}{p}$$?

Answer 1

Mertens's formula says that $$ \prod_{\substack{p\le z \\ p \text{ prime}}} \frac{p-1}p \sim \frac1{e^\gamma\log z}, $$ where $\gamma$ is Euler's constant. (Here, $\sim$ means that the limit of the quotient of the two expressions equals $1$.) In particular, $$ \prod_{i\le y} \frac{p_i-1}{p_i} = \prod_{\substack{p\le p_y \\ p \text{ prime}}} \frac{p-1}p \sim \frac1{e^\gamma\log p_y}. $$ Since $p_y\sim y\log y$ by the prime number theorem, and $\log(y\log y) \sim \log y$, this last formula becomes $$ \prod_{i\le y} \frac{p_i-1}{p_i} \sim \frac1{e^\gamma\log y}. $$ These results give you what you need to deduce your equality.