TITLE: The limit of the following product? What is the closed form of the value?
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.