Does $\Pi_{i=1}^{\infty}(\frac{p_{i}-1}{p_{i}})$ converges?

Question

Does $\prod_{i=1}^{\infty}\left(\frac{p_{i}-1}{p_{i}}\right)$ converges?

My motivation to problem is:

Let $n \in {2,3,4,5,6,...}$ and $n=p_{1}^{l_{1}}p_{2}^{l_{2}}\ldots p_{k}^{l_{k}}$

$$\phi(n)=n\prod_{i=1}^{k}\left(\frac{p_{i}-1}{p_{i}}\right)$$

I want to see that as the number $k$ of primes in factorization of $n$ approaches $\infty$, then $\phi(n)$ approaches $nT$, where $T=\prod_{i=1}^{\infty}(\frac{p_{i}-1}{p_{i}})$ if the product converges.

But I think it diverges since this product is convergent iff the series of reciprocals of primes is convergent. (Here I use the theorem that: Let $x_{n} \geq 0$,then: $\prod_{n=1}^{\infty}(1-x_{n})$ converges iff $\sum_{n=1}^{\infty}(x_{n})$ converges).

Am I true?

My desire to find continuous function or polynomial approximation that has values near to $\phi$ as the number gets larger.

Answer 1

No! The product diverges to $0$. Because the series $\sum\frac1{p_n}$ does not converge.

Answer 2

Let us define$$a_n=\prod_{i=1}^n\frac{p_i-1}{p_i}\implies\frac{1}{a_n}=\prod_{i=1}^n\frac{1}{1-p_i^{-1}}=\prod_{i=1}^n\left(1+\frac{1}{p_i}+\frac{1}{p_i^2}+\cdots\right)$$Therefore, $1/a_n$ equals the sum of the reciprocals of all the positive integers whose prime factors are among $p_1,p_2,\ldots,p_n$. As $n$ tends to infinity, this diverges, hence $a_n$ tends to $0$. Therefore, $$\prod_{i=1}^\infty\frac{p_i-1}{p_i}=0$$so indeed the product diverges.