What is the range of $a $ such that $\frac{\log N}{N^{3a+1/2}}\rightarrow 0$ as $N\rightarrow \infty$?

Question

I want to find the range of $a$ such that $\frac{\log N}{N^{3a+1/2}}\rightarrow 0$ as $N\rightarrow\infty$. The answer to this question hinges on conditions under which $logN$ diverges slower than the numerator, which is a power function of $N$.

Is it true that $\log N$ diverges slower than $N^\gamma$ for any $\gamma>0$ (so that the range for $a$ is $a>-1/6$)?

Yes, it is true: for any $\alpha,\beta>0$, $:\lim_{x\to\infty}\dfrac{\log^\alpha x}{x^\beta}=0$ is a basic limit. — Bernard, Oct 07 '20 at 13:13

Student · Accepted Answer · 2020-10-07T19:06:04.603

2

Yes, $\lim_{N \to \infty} \frac{\log N}{N^{3a+1/2}} \to 0 \iff \Big((3a+\frac{1}{2})>0 \implies a>\frac{-1}{6}\Big)$.

What follows is not a proof but you can play with a few values on Wolfram Alpha: https://www.wolframalpha.com/input/?i=limit. For example, enter $\frac{\ln x}{x^{10^{-1000}}}$ as the function and "infinity" as the value to approach on the Wolfram Alpha page, and you will see the fraction converge to $0$.

edited Oct 07 '20 at 19:06

answered Oct 07 '20 at 05:30

Student

136

Thank you very much! This is very helpful! – ExcitedSnail Oct 07 '20 at 05:36

score 2 · Answer 2 · answered Oct 07 '20 at 05:36

2

See the following picture, which is found in https://kconrad.math.uconn.edu/blurbs/analysis/growth.pdf.

answered Oct 07 '20 at 05:36

ExcitedSnail

503

Can you sumarise the link in a few sentences? A link with a picture is not that helpful, as pictures cannot be read by visually impaired people with screen-readers and search engines. – Toby Mak Oct 22 '20 at 11:35

What is the range of $a $ such that $\frac{\log N}{N^{3a+1/2}}\rightarrow 0$ as $N\rightarrow \infty$?

2 Answers2