Is this true that $n\log\left(\frac{p_n}{p_{n+1}}\right)$ is bounded, where $p_n$ is the $n$-th prime number?
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what do you mean by $p_n$ is prime number $n$? – Jack Yoon May 16 '14 at 14:39
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I mean $p_n$ is the nth prime number. – Papagon May 16 '14 at 14:44
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1Try $p_n \sim n \log n$. – lhf May 16 '14 at 14:53
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1Also $\frac{p_{n+1}}{p_n} \to 1$, but this is probably overkill. – lhf May 16 '14 at 14:56
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@lhf: $p_n=n\log(n)+O(n\log(\log(n)))$. Trying to estimate the gap between primes using $n\log(n)$ seems too weak here. – robjohn May 16 '14 at 15:50
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Seems unbounded:
Let $g_n = p_{n+1} - p_n$ be the prime gap, then Westzynthius's result (see link below) states that $\lim\sup \left[ g_n/(\log p_n) \right] = \infty$, hence
$$\lim \sup n \log(p_{n+1}/p_n) = \lim \sup n \log (1 + g_n/p_n) = \lim \sup n g_n/ p _n = \lim \sup g_n/\log n = \infty$$
PA6OTA
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@DietrichBurde: they are negatives of each other. If one is bounded, so is the other. – robjohn May 16 '14 at 15:45
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1@PA6OTA: although still a conjecture, this is probably as close as we can get now. (+1) – robjohn May 16 '14 at 15:47
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$\frac{p_n}{p_{n+1}}<1$ for all $n$, so when $n\to\infty$, $(\frac{p_n}{p_{n+1}})^n\to0$. so $$n\log(\frac{p_n}{p_{n+1}})=\log(\frac{p_n}{p_{n+1}})^n\to-\infty$$ so this is not bounded
Martial
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1And also because this is false reasoning... if $x_n \in [0,1[$ satisfies $x_n \to x \in [0,1[$, then $y_n = \sqrt[n]{x_n}$ satisfies $0 \le y_n < 1$ and $y_n^n \to x$. So this reasoning is not sufficient to show $(p_n/p_{n+1})^n \to 0$. – Patrick Da Silva May 16 '14 at 15:05