3
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.