Question

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.

Answer 1

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.