First of all, sorry for my ignorance. I'm just a common guy interested in maths.
I have read a lot about the product $$\prod^{N} \frac{1}{1-\frac{1}{p}}=\prod^{N} \frac{p}{p-1}$$ (Where $p$ is a prime number); including its behaviour and its relationship with Zeta function and the Harmonic series when $N \to \infty$.
But what's about this other product? $$\prod^{N} \frac{p-1}{p-2}$$ Is it any lower or upper bound known for it? What is its asymptotic behaviour? Where could I find information about it?
Thank you!
I assume that, for this product, you don't want to include $2$ as one of your primes, so I'll consider
$$\prod_{2<p\leq N} \frac{p-1}{p-2}$$
Note that
$$\frac{n}{n-1}$$
is a strictly decreasing function of $n$, so we have that
$$\frac{p_{n-1}}{p_{n-1}-1} \geq \frac{p_n-1}{p_n-2} > \frac{p_n}{p_n-1}$$
where $p_n$ is the $n$th prime ($p_1=2,p_2=3,\cdots$)
So, the product can be bounded by
$$\prod_{n=2}^{\pi(N)} \frac{p_{n-1}}{p_{n-1}-1} \geq \prod_{n=2}^{\pi(N)} \frac{p_n-1}{p_n-2} > \prod_{n=2}^{\pi(N)} \frac{p_n}{p_n-1} $$
Denote $X_k = \prod_{n=1}^{k} p_k$ as the $k$th primorial, and let $\phi(n)$ represent the Euler totient function. We can now bound our product by
$$\frac{X_{\pi(N)-1}}{\phi\left(X_{\pi(N)-1}\right)} \geq \prod_{2<p\leq N} \frac{p_n-1}{p_n-2} > \frac{X_{\pi(N)}}{2\phi\left(X_{\pi(N)}\right)}$$
We can use Mertens' third theorem to give that
$$\frac{X_{\pi(N)}}{\phi\left(X_{\pi(N)}\right)}\sim e^{\gamma} \log N$$
So, the product is asymptotically between $\frac{e^{\gamma}}{2}\log N$ and $e^{\gamma}\log N$, so it is asymptotically logarithmic.
Getting a very precise upper bound is extremely difficult, but we can do it assuming the Riemann Hypothesis using "RH Equivalence 3.25" here.