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## On a limit involving consecutive primes

I observed the following result empirically based on numerical evidence.

Conjecture: Let $[x]$ denote the greatest integer function. If $a_n$ is an strictly increasing sequence of positive real numbers such that $a_1 \ge 1$ and $\lim_{n \to \infty}\frac{1}{a_n}=0$ and $\lim_{n \to \infty}\frac{a_{[nt]}}{a_n} = t$, for every real $0 \le t < 1$. I observe empirically that $$\lim_{n \to \infty} \frac{\ln\Big(\frac{\sqrt[n]{a_n}}{\sqrt[n+1]{a_{n+1}}}-1\Big)}{\ln n} = -2.$$

Motivation: The sequence of natural numbers $n$, prime numbers $p_n$ etc. I am particularly interested in the limit involving the sequence of primes $p_n$. This is a generalization of the Firoozbakht conjecture and would imply Cramer's conjecture. The very fact Cramer's conjecture pops up is enough to discourage any attempt of proof. Nonetheless, I shall raise this as an open question.

Question: More than the main conjecture, I am interested in the following special case question. Prove or disprove that

$$\frac{\ln\Big(\frac{\sqrt[n]{p_n}}{\sqrt[n+1]{p_{n+1}}}-1\Big)}{\ln n} = -2 + \frac{\ln\ln n}{\ln n} + O\Big(+ \frac{\ln\ln n}{\ln^2 n}\Big).$$

Edit: Added the asymptotic form of the limit for primes.

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For some related discussion, see mathoverflow.net/questions/90327/… – Pat Devlin Apr 10 2012 at 13:30
The conjecture holds for a_n = C*n, where C is a positive constant. – Pat Devlin Apr 10 2012 at 13:36
It is very unclear to me why you change (for two questions) the existing meaningful titles to the same (current) one that hardly conveys any information. – quid May 30 at 11:55
If you feel the need to change the title of a question, you could at least turn it into a question. – S. Carnahan May 30 at 12:05

The general conjecture shouldn't hold. To begin with, the assumptions do not even seem to imply that $\sqrt[n]{a_n}>\sqrt[n+1]{a_{n+1}}$ for all sufficiently large $n$. For a more sophisticated example related to the prime counting function, let $a_n=\psi(n)$ be the Chebyshev function. (Make it strictly increasing by adding some tiny amount.) Assuming Riemann hypothesis, $a_n=n+O(\sqrt n(\log n)^2)$. If I have done my calculations right, this implies $$\frac{\sqrt[n]{a_n}}{\sqrt[n+1]{a_{n+1}}}-1=\frac{\log n+a_n-a_{n+1}}{n^2}+O\left(\frac{(\log n)^6}{n^{5/2}}\right).$$ In particular, if $n=p-1$ where $p$ is prime, then $a_{n+1}=a_n+\log(n+1)$, hence the first term on the right disappears, and $$\frac{\log\left(\frac{\sqrt[n]{a_n}}{\sqrt[n+1]{a_{n+1}}}-1\right)}{\log n}\le-2.5+o(1).$$ This happens for infinitely many $n$, hence the limit is not $-2$. I didn't try it with the $p_n$ function, but I have doubts.
 I should also say I'm pretty sure one can cook up an artificial counterexample to the general conjecture by hand, without using RH and such stuff. The reason I'm using the Chebyshev function is to make it close in spirit to the $p_n$ sequence. – Emil Jeřábek Apr 10 2012 at 14:33 (For an explicit elementary counterexample, $a_n=n+\sum\{t\in\mathbb N:e^t\le n\}$ should work. The idea is that it is quite close to $n$, but there are jumps of size $\log n$ now and then.) – Emil Jeřábek Apr 10 2012 at 14:49 @Emil, Thanks for the inputs. I derived an asymptotic expansion for primes, based on assumptions that we can use Cipolla's asymptotic expansion of the n-th prime. I have presented the limit asked in teh question for primes in an asymptotic format. I guess the -2 limit is not general but in case of primes, both calculations and the asymptotic series both suggest the limit of -2. But I doubt is, are we justified to use Cipolla's expansion to evaluate this limit?? – dx Apr 11 2012 at 13:37 @Neel: No, you can't use any approximation. The limit is likely to be $-2$ for any sufficiently "smooth" function, and that's exactly what the approximation provides. The examples above show that the limit may fail to be $-2$ if there are local irregularities; in both examples, $a_n$ is fairly well approximated (with much smaller relative error than Cipolla) by the sequence $a_n=n$, for which the limit is $-2$. – Emil Jeřábek Apr 11 2012 at 16:08