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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.

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    $\begingroup$ Looks like it's tending to $\infty$ to me (e.g. the product is 12.63 for $k = 10^3$, and 35.12 for $k = 10^4$). I still think it's a valid question to determine the asymptotic behaviour of this product; it seems to grow a bit faster than $\log k$. $\endgroup$
    – David Loeffler
    Commented Nov 6, 2015 at 10:27
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    $\begingroup$ The infinite product itself goes to zero. With $k$ on there is goes to infinity. So what factor should be used to get something interesting? $\endgroup$
    – Gerald Edgar
    Commented Nov 6, 2015 at 15:12
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    $\begingroup$ I believe the correct order of growth of the product is $(k \ln k)^{-1/2}$, up to some multiplicative constant. See my answer. It shows $\max\{ \prod_{n=1}^k \frac{P_{2n-1}}{P_{2n}} , \prod_{n=1}^k \frac{P_{2n}}{P_{2n+1}} \} \gg (k \ln k)^{-1/2}$. $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 6, 2015 at 15:26
  • $\begingroup$ @GeraldEdgar I was kind of interested in that question, I thought that k may get something. interesting $\endgroup$
    – user82419
    Commented Nov 6, 2015 at 15:45

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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.

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  • $\begingroup$ Thanks for pointing for a glitch in my answer. I'm taking it down to see if I can fix it. But your answer is flawed as well: the closeness of $F_k$ and $G_k$ (within a bounded factor) is unclear at all. E.g., it may be plausible that they are close with a factor of $\log k$ or $\sqrt{\log k}$ or so. $\endgroup$
    – Max Alekseyev
    Commented Nov 6, 2015 at 15:32
  • $\begingroup$ Indeed, I didn't provide a proof of that, basically because it might be an extremely hard problem (which is equivalent to the original question, as follows from my answer). But intuitively at least I believe this assertion. (Also, being off by a polylogarithmic factor doesn't change the conclusion.) $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 6, 2015 at 15:35

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