Let's say I want to calculate: $$\lim_{n \to \infty} \frac{\prod_{k=2}^{n}{p_{k}*p_{k}}}{\prod_{k=2}^{n}({p_{k}-2)*p_{k+1}}}$$ where $p_{k}$ is the $k^{th}$ prime.
Basically this is $(3*3)/(1*5) * (5*5)/(3*7) *(7*7)/(5*11) {...}$
Is this infinite product equal to $0$? I am assuming it is because $$\frac{{p_{k}*p_{k}}}{({p_{k}-2)*p_{k+1}}}$$ is mostly less than 1 for large primes, but I cannot find a good proof of why the infinite product should converge to $0$ rather than some other finite number.
The product
$$\prod_{k=2}^n{\frac{p_k \cdot p_k}{(p_k - 2)\cdot p_{k+1}}}$$
does in fact tend to $0$ as $n \to \infty$.
We can show this by splitting it up into two products as shown below
$$\prod_{k=2}^n{\frac{p_k}{p_{k+1}}} \cdot \prod_{k=2}^n{\frac{p_k}{p_k - 2}}$$
The terms on the left telescope, so we're left with
$$\frac{3}{p_{n+1}} \cdot \prod_{k=2}^n{\frac{p_k}{p_k - 2}} = \frac{3}{p_{n+1}} \cdot \prod_{k=2}^n{\left(1 + \frac{2}{p_k - 2}\right)}$$
As $n$ tends to infinity,
$$\frac{1}{p_{n+1}} \to \frac{1}{n\log(n)}, \ \ \text{ by the PNT, and}$$
$$\prod_{k=2}^n{\left(1 + \frac{2}{p_k - 2}\right)} = O(\log^2(n)).$$
When you multiply the two, their product clearly approaches zero.