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Let $p_n$ be the $n$th prime, $n\ge 1$. I was wondering if the following product converges $$P=\prod _{k=1}^{\infty} \frac{p_{k+1}}{\sqrt{p_k p_{k+2}}}$$

I have computed some numeric examples and the result $1.224$ keeps the same $3$ decimals when $k$ goes from $10000$ on, which makes me think it converges, but I don't know how to prove or disprove it.

Any idea?

Thank you for your attention

Edit

The result here that $$\lim_{n\to\infty}\frac{p_{n+1}}{p_n}=1$$ and the comment made by Donald Splutterwit that the general term of the product can be written as $$\sqrt{\frac{3}{2}}\sqrt{\frac{p_n}{p_{n+1}}}$$ let me conclude that the product converges to $$P=\sqrt{\frac{3}{2}}$$

Raffaele
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    This product from $k=1$ to $n$ is equal to $\sqrt{\frac{p_2}{p_1}}\sqrt{\frac{p_{n+1}}{p_{n+2}}}$, right? – NN2 Feb 13 '21 at 21:31
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    Telescope ... Converges to $\sqrt{\frac{3}{2}}$. – Donald Splutterwit Feb 13 '21 at 21:50
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    @DonaldSplutterwit: you have proof that $\frac{p_{n}}{p_{n+1}}$ converges to $1$? – NN2 Feb 13 '21 at 22:02
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    @NN2 Yeah, I think you are right ... that result is stated here https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds ... maybe you could answer this for Raffaele & keep it out of the unanswered list ? – Donald Splutterwit Feb 13 '21 at 22:10

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