Find $\lim_{n\rightarrow \infty}\frac{(\log n)^a} { n^b}$ when $ a,b >0$.

Question

I need to prove that $(\log n)^a$ will always be smaller than $n^b$ as $n$ get larger (to infinity), also the condition is $a, b > 0$ ($a,b$ are random). I test the graph and it's true but I cant' prove it. I can prove that $\log(n)$ always $< n$, $\log(\sqrt{n}) < \sqrt{n}$ but not the title.

Sorry, I type math on Word but when copying here it didn't work so I plain typing the math function.

Answer 1

Numerator:

$(\log n) ^a=(\log ((n^{b/a})^{a/b}))^a=$

$((a/b) \log n^{b/a})^a=$

$(a/b)^a (\log n^{b/a}) ^a$

Denominator:

$(n^{b/a})^a$

Putting together:

$(a/b)^a \left (\dfrac {\log n^{b/a}} {n^{b/a}} \right) ^a $

With $k:=(a/b)^a$ and $z:=n^{b/a}$ we have:

$k \left (\dfrac{\log z}{z} \right)^a. $

Take the limit $z \rightarrow \infty. $

Answer 2

Note that: \begin{equation} \frac{\log(n)^{a}}{n^{b}}=\left(\frac{\log(n)}{n^{\frac{b}{a}}}\right)^{a} \end{equation} and consider the function $f(x)=\frac{\log(x)}{x^{\frac{b}{a}}}$ for $x>0$. It follows from L'hopital's rule that \begin{equation} \lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\frac{1}{x\ln(10)}}{\frac{b}{a}x^{\frac{b}{a}-1}}=\lim_{x\to\infty}\frac{a}{b\ln(10)x^{\frac{b}{a}}}=0 \end{equation} and because $x\mapsto x^{a}$ is continuous at $x=0$, it then follows that $\lim_{n\to\infty}\frac{\log(n)^{a}}{n^{b}}=0$ as you intuited with what you were trying to prove.

Answer 3

Since you've established that $\log x < x$ for $x > 0$ you can note that $r \log x = \log x^r < x^r$ for any exponent $r$, and thus $$\log x < \frac 1r x^r$$ whenever $x$ and $r$ are both positive. If $a > 0$ and $x \ge 1$ you get $$0 \le (\log x)^a < \frac 1{r^a} x^{ra}$$ and then for any $b$ $$0 \le \frac{(\log x)^a}{x^b} < \frac 1{r^a} x^{ra-b}.$$ Provided that $b > 0$ you can select $r > 0$ small enough that $ra - b < 0$, and then use the squeeze theorem.