Proving $\lim_{n\to \infty} \frac{(\ln n)^a}{n^b} = 0$ for all $a,b>0$

Question

How can I prove this:

$$\lim_{n\to \infty} \frac{(\ln n)^a}{n^b} = 0 \quad \forall\, a,b > 0$$

Any ideas or tips? I tried to use L'Hôpital's rule but that led into nothing.

This question must be asked at least one time here... – Yes Oct 11 '15 at 15:40 — Yes, Oct 11 '15 at 15:40

score 1 · Accepted Answer · edited Nov 15 '23 at 20:21

1

You can write that

$$\frac{(\ln(n))^a}{n^b} = \left( \frac{\ln(n)}{n^{\frac{b}{a}}} \right)^a$$

Then you want to calculate $\lim_{n\to \infty} \frac{\ln(n)}{n^c}$ for $c>0$ (here Hospital's rule works well). And you conclude by continuity of $x\mapsto x^a$

edited Nov 15 '23 at 20:21

user577215664

40,625

answered Oct 11 '15 at 15:34

Tryss

14,310

You don't need continuity, just a strictly increasing transformation. See my answer. – user2566092 Oct 11 '15 at 15:37
Okay, I understand what you are doing here...so you try to say that: lim(x^a) = lim(x)^a? – 今天春天 Oct 11 '15 at 16:26
Thanks, now I understand...this did it for me. – 今天春天 Oct 11 '15 at 16:37

score 0 · Answer 2 · answered Oct 11 '15 at 15:34

0

Take the $1/a$th root of both top and bottom. That is a strictly increasing transformation so if the limit is $\infty$ then the original limit is $\infty$ as well. Now you can use L'Hopital and get something.

The new limit you will get is $\lim_{x \to \infty}\frac{\ln x}{x^{b-a}}$ which you can evaluate.

answered Oct 11 '15 at 15:34

user2566092

26,142

But what if b = 3 and a = 4? Then the limit is not 0... – 今天春天 Oct 11 '15 at 16:20

Proving $\lim_{n\to \infty} \frac{(\ln n)^a}{n^b} = 0$ for all $a,b>0$

2 Answers2

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