This re-asks a question I posed on MSE:

Q. Does this infinite product converge?

$$ \frac{2}{3}\cdot\frac{7}{5}\cdot\frac{11}{13}\cdot\frac{19}{17}\cdot\frac{23}{29}\cdot\frac{37}{31} \cdot \cdots \;. $$ I call this the primes snake-product:

Out to primes of size $10^{10}$, MSE user @Peter calculated the product to be $\approx 0.9048$.

@Wojowu showed that the question is likely difficult, relying on estimates of alternating sums of prime gaps, and that perhaps convergence is beyond current knowledge. I re-pose the question to see if indeed this is the case, or might known bounds suffice to establish convergence.

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    $\begingroup$ Taking the log and using Bertrand's postulate should give rise to a series whose convergence can be studied more easily. $\endgroup$ Feb 5, 2019 at 14:27
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    $\begingroup$ I think you should expect it to converge to zero. The terms alternate between being greater than 1 and less than 1. Taking them in pairs, and considering them as”random variables”, the $n$th random variable should be something like $1\pm 1/\log n$. In similar situations for martingales, such a product should converge almost surely to 0. $\endgroup$ Feb 6, 2019 at 1:06
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    $\begingroup$ @AnthonyQuas How did you arrive at $1\pm 1/\log n$? Naively, the gap between $p_n$ and $p_{n+1}$ is on average $\log p_n\approx\log n$, so $\frac{p_n}{p_{n+1}}\frac{p_{n+3}}{p_{n+2}}\approx\frac{1}{1+a}\frac{1+a+b+c}{1+a+b}\approx 1+a-c$ where $a,b,c\approx\log n/p_n\approx 1/n$. With $1\pm 1/n$, the product should converge. $\endgroup$
    – Wojowu
    Feb 6, 2019 at 15:41
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    $\begingroup$ Thanks @Wojowu: this was a mental arithmetic error on my part (and your corrected version fits the observed values better). $\endgroup$ Feb 7, 2019 at 5:38
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    $\begingroup$ @Wojowu: $1+a-c$ or $1-a+c$? $\endgroup$ Feb 11, 2019 at 18:41


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