Define $s_n=p_{n+1}-p_n$, where $p_n$ is the $n$th prime number, now how to show that $\lim_{n \rightarrow \infty} \inf \frac{s_n}{\log n} \leq 1$ I used the result from the prime number theorem: $p_n \sim n \log n$, but have a difficult time dealing with the $\inf$ part.
How to prove this inequality using prime number theorem
2 Answers
Here we go: $p_n \sim n\log n$, and $p_{2n}\sim 2n\log(2n)\sim 2n\log n$. So $p_{2n}-p_n=(p_{2n}-p_{2n-1})+(p_{2n-1}-p_{2n-2})+\cdots +(p_{n+2}-p_{n+1})+(p_{n+1}-p_{n})\sim n\log n.$ As there are $n$ terms and the average is $\log n$, one needs to be smaller then $\log n$, which means lim inf is smaller.
It should not be too hard to make this sketch rigorous.
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0That's pretty clever. – 2011-10-01
To deal with the infimum, remember that $p_n \sim n \log n + o(n \log n)$, see this page for more accurate expansion.
The sequence $(p_{n+1}-p_{n})/\log n$ has lots of spikes. The infimum of this sequence would be the lower envelope for it. For sufficiently large $n$ $ \begin{eqnarray} \inf \frac{p_{n+1} -p_n}{\log n} &<& \frac{(n+1) \log(n+1) - n \log n}{\log n}= \frac{(n+1) \left(\log n + \log(1+\frac{1}{n})\right) - n \log n}{\log n}\\ &=& 1 + \frac{n+1}{\log n} \log\left( 1 + \frac{1}{n} \right) < 1 + \frac{n+1}{\log n} \frac{1}{n} = 1 + o(1) \end{eqnarray} $
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1The first inequality, which is the most important, lacks justification. – 2011-10-01