Let $\{a_n\}\subset N $ be an increasing positive integer sequence, i.e., $ 0<a_1<a_2<\cdots<a_n<a_{n+1}<\cdots.$ Assume that $$\lim\limits_{n\to \infty}\frac{\log n}{\log a_n}=1,$$ can we have $\lim\limits_{n\to \infty}\frac{a_n}{a_{n+1}}=1$? Thanks!


Set $t_m=3^{3^m}$, then define $a_n=3^mn $ when $3^{3^m}\leq n<3^{3^{m+1}}$

Then $n\leq a_n \leq \frac{n\log n}{\log 3} \leq n \log n$, so $$ \frac{\log n}{ \log \log n+\log n} \leq \frac{\log n}{\log a_n} \leq1 $$

Therefore $$\frac{\log n}{\log a_n}\to 1$$

But $$ \frac{a_{3^{3^m}}}{a_{3^{3^m}-1}} = \frac{3^m\cdot 3^{3^m}}{3^{m-1}\cdot (3^{3^m}-1)}\to 3$$

  • $\begingroup$ I am confused by $m$. What is $m$? $\endgroup$ Mar 15 '19 at 1:25
  • $\begingroup$ @JackyChong We define the sequence $t_m=3^{3^m}$, and then we define $a_n=3^m n$ where $m$ is the unique $m$ such that $t_m \leq n < t_{m+1}$. $\endgroup$
    – clark
    Mar 15 '19 at 1:28
  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – ljjpfx
    Mar 15 '19 at 1:29

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