$\lim_{n\to\infty} \frac{(2-1)\times (3-1) \times (5-1) \times \dots \times (p_n-1)}{2\times 3 \times 5 \times \dots \times p_n}$

Question

What is this limit

$$\lim_{n\to\infty} \frac{(2-1)\times (3-1) \times (5-1) \times \dots \times (p_n-1)}{2\times 3 \times 5 \times \dots \times p_n}$$

Here, $p_n$ is the $n$-th prime.

I figured that it is $\lim_{n\to\infty}\frac{\Phi(n)}{n}$ for some specific $n$'s, where $\Phi(n)$ is Euler's totient function, but how do I even begin to estimate something like this?

Thanks

Answer 1

$e^x\ge 1+x$ for all $x$. So $$ \prod_{p_n \le x}\left(1-\frac{1}{p_n}\right) \le \mathrm{exp}\left(-\sum_{p_n\le x}\frac{1}{p_n}\right)= \mathrm{exp}\left(-\log \log x\,(1+o(1))\right) \to 0. $$

Answer 2

Note that

$$\frac{(2-1)\times (3-1) \times (5-1) \times \dots \times (p_n-1)}{2\times 3 \times 5 \times \dots \times p_n}=\prod_{k=1}^n \left(1-\frac1{p_k}\right)=\frac{1}{\prod_{k=1}^n \frac1{1-p_k^{-1}}}$$

and since by Euler product identity

$$\prod_{p} \frac1{1-p_k^{-1}}=\sum_{n=1}^\infty \frac1n=\infty$$

we have that

$$\lim_{n\to\infty} \,\prod_{k=1}^n \left(1-\frac1{p_k}\right)=\frac1{\infty}=0$$