Assume that $P\_n$ is the $n$'th prime: Please help me solve the following $$\lim\_{k\to\infty} {k}\prod\_{n=1}^k \frac{P\_{2n-1}}{P\_{2n}}$$


I am not really sure quite where to start here as I am dealing with primes, one of my weaknesses I want to make the important point: I am not a professional mathematician, in fact only a high school student, but am desperate to learn all the info I can. So if it would be possible, for me, could you please namedrop theorems in your answer? I also want to apologize if it is rude to ask questions here not being remotely professional.


Back to the problem, however, I did note that the values do seem to converge. If they do converge, if possible, find a closed form. If there is no closed form, simply ignore my last sentence.


EDIT: if there is a factor that generates a convergent product, please let me know, as k alone doesn't do it, I would want to know what does.



The limit of the following product? What is the closed form of the value?
*Clarification:* This is not a full answer to the original question (which might be a very hard one), but rather an heuristic argument.


Let $F\_k = \prod\_{n=1}^{k} \frac{p\_{2n-1}}{p\_{2n}}, G\_k = \prod\_{n=1}^{k} \frac{p\_{2n}}{p\_{2n+1}}$. You are interested in $F\_k$, but heuristically at least, they should be close to each other, although it might be hard to prove:
$$ F\_k \sim G\_k$$
Fortunately, $F\_k \cdot G\_k$ has a nice form - it is $\prod\_{i=1}^{2k} \frac{p\_i}{p\_{i+1}}$, which telescopes to $\frac{p\_1}{p\_{2k+1}}$. By the prime number theorem,


$$p\_{2k+1} \sim (2k) \log k$$


Hence we'd expect 
$$F\_k \sim \sqrt{F\_k \cdot G\_k} \sim \sqrt{\frac{2}{2k \log k}} = \sqrt{\frac{1}{k \log k}}$$
This implies $kF\_k \sim \sqrt{\frac{k}{\log k}}$, which goes to infinity as $k$ goes to infinity.


