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The general rule is discussed here but that doesn't solve my problem. I want to prove that $$\lim_{n \to \infty} \dfrac{e^{c \sqrt{\ln n . \ln \ln n}}}{n^{\epsilon}}=0$$ where $c>0$ a fixed constant and $\epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :

$\lim_{n \to \infty} \dfrac{(\ln n . \ln \ln n)^k}{n^{\epsilon}}=0$ for any $k \in \mathbb{N}$. Does it imply $\lim_{n \to \infty} \dfrac{e^{c \sqrt{\ln n . \ln \ln n}}}{n^{\epsilon}}=0$?

If not, any hint for solving $\lim_{n \to \infty} \dfrac{e^{c \sqrt{\ln n . \ln \ln n}}}{n^{\epsilon}}=0$ would be appreciated.

2 Answers2

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The statement is equivalent to $c\sqrt {\ln\, n \ln\ln \, n}-\epsilon \ln\, n \to -\infty$. To show this write this as $-\ln\, n [\frac {c\sqrt {\ln\, n \ln\ln \, n}} {-\ln \, n} +\epsilon]$. Can you show that the expression inside [] goes to $\epsilon$? It amounts to showing that $\frac {\ln \ln \,n} {\ln\, n} \to 0$ for which you can use L'Hopital's Rule.

Kavi Rama Murthy
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$$\displaystyle L=\lim_{n\to\infty}\frac{e^{c\sqrt{\ln n\cdot\ln\ln n}}}{n^{\epsilon}}$$

$\displaystyle\ln L=\lim_{n\to\infty}c\sqrt{\ln n\cdot\ln\ln n}-\epsilon\ln n$

$\displaystyle=\lim_{x\to\infty}c\sqrt{x\cdot\ln x}-\epsilon x$

$\displaystyle=\lim_{x\to\infty}\frac{c\sqrt{\frac{\ln x}x}-\epsilon}{\frac1x}\to-\infty\ \ \ \because\lim_{x\to\infty}\frac{\ln x}x=0 $

$\therefore L\to e^{-\infty}=0$

Shubham Johri
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