Prove that $\lim_{n \to \infty} \frac{(\ln{n})^2}{\sqrt{n}} = 0$

Question

Prove that $\displaystyle \lim_{n\to\infty}\frac{(\ln{n})^2}{\sqrt{n}}=0$. In general, is it true that $\displaystyle\lim_{n\to\infty}\frac{(\ln{n})^k}{\sqrt{n}}=0$ for all power of $k$?

Answer 1

Change this to functions and use l'Hospital's Rule:

$$\lim_{x\to\infty}\frac{\log^k x}{\sqrt x}=\lim_{x\to\infty}\frac{k\log^{k-1}x}{2\sqrt x}=\overbrace{...}^{k-1\;\text{times}}=0$$

Answer 2

We do not need to apply L'Hôpital's rule $k$ times if we use continuity. For $p>0$ and $q>0$, we have that $$ \lim_{x\to\infty}\frac{\log^px}{x^q} =\lim_{x\to\infty}\Bigl(\frac{\log x}{x^{q/p}}\Bigr)^p =\Bigl(\lim_{x\to\infty}\frac{\log x}{x^{q/p}}\Bigr)^p =\Bigl(\lim_{x\to\infty}\frac1{(q/p)x^{q/p}}\Bigr)^p=0 $$ using continuity and l'Hôpital's rule.

Answer 3

For all $p,q>0$,

$$\lim_{n\to\infty}\frac{\ln^p(n)}{n^{q}}=\lim_{{n^{p/q}}\to\infty}\frac{\ln^p(n^{p/q})}{(n^{p/q})^q}=\left(\frac pq\right)^p\lim_{{n}\to\infty}\frac{\ln^p(n)}{n^p}=\left(\frac pq\right)^p\lim_{{n}\to\infty}\left(\frac{\ln(n)}{n}\right)^p.$$

So there is no need to worry about the exponents and it suffices to settle the case of $$\lim_{{n}\to\infty}\frac{\ln(n)}{n}=\lim_{e^{n}\to\infty}\frac{n}{e^n}\le\lim_{n\to\infty}\frac{n}{1+n+\dfrac{n^2}2}.$$

Answer 4

Maybe this will help you understand it better:

Make a change of variable:

$$\ln n=p$$

$$\lim_{n \to \infty} \frac{(\ln n)^2}{\sqrt{n}}=\lim_{p \to \infty} \frac{p^2}{e^{p/2}}$$

Exponent contains every positive power of $p$ in its series expansion. (If you have not studied Taylor series yet, the limit definition gives the same result).

In this case:

$$e^{p/2}=1+\frac{p}{2}+\frac{p^2}{8}+\frac{p^3}{48}+\cdots+\frac{p^k}{2^k k!}+\cdots$$

So, basically, you can write:

$$e^{p/2} \geq 1+\frac{p}{2}+\frac{p^2}{8}+\frac{p^3}{48}\geq \frac{p^3}{48}$$

$$\lim_{p \to \infty} \frac{p^2}{e^{p/2}} \leq \lim_{p \to \infty} \frac{p^2}{\frac{p^3}{48}}=\lim_{p \to \infty} \frac{48}{p}=0$$

For any positive powers $k$ and $m$ there will always be infinite amount of terms in exponential function $e^{mp}$ with powers higher than k, thus making the limit go to zero.

Answer 5

First, a basic lemma about real polynomials: Let $p(x)=\sum_{j=0}^n a_jx^j$ with $n>0$ and $a_n\ne 0.$ For all sufficiently large $|x|$ we have $|p(x)|>|a_n x^n/2|.$ Proof: When $|x|>1$ and $|x|>\max_{j<n}2 n |a_j a_n^{-1}|$ we have $|a_jx^j|\leq|a_jx^n|/|x|<|a_jx^n/(2 n |a_j a_n^{-1}|)=|a_n x^n|/2 n|$ for $j<n,$ so $$|p(x)|\geq |a_nx^n|-\sum_{j=0}^{n-1}|a_jx^j|>|a_nx^n|(\;1-\sum_{j=0}^{n-1}1/2 n\;)=|a_nx^n/2|.$$

Second, $\lim_{x\to \infty} \log x=\infty.$ Let $f(x)$ be the largest integer not exceeding $\log x.$ So $f(x)\to \infty$ as $x\to \infty.$

Take any positive integer $k.$ When $f(x)> k$ we have $$x=e^{\log x}\geq e^{f(x)}=(\;1+(e-1)\;)^{f(x)}=$$ $$=\sum_{j=0}^{f(x)}(e-1)^j\binom {f(x)}{j}>(e-1)^k\binom {f(x)}{k}.$$

Treating this last expression as a polynomial of degree $k$ in the variable $f(x),$ by the lemma we have $$(e-1)^k \binom {f(x)}{k}>(e-1)^k f(x)^k/(k! 2)$$ for all sufficiently large $f(x),$ which is for all sufficiently large $x.$

Therefore $x>(e-1)^k f(x)^k/(k! 2)$ for sufficiently large $x.$ This gives $$\log x<1+f(x)<1+A(k)x^{1/k}$$ where $A(k)=(k! 2)^{1/k}/(e-1).$ (Again, only for all sufficiently large $x.$)

Therefore for any $r>0$ we may take $k$ large enough that $1/k<r$ and we have $$(\log x)/x^r<(1+A(k)x^{1/k})/x^r$$ for all large $x.$ So in conclusion, $$r>0\implies \lim_{x\to \infty} (\log x)/x^r=0.$$

And for any $k>0$ we have $(\log x)^k/\sqrt x=( \;(\log x)/x^{1/2 k}\;)^k$ which also $\to 0$ as $x\to \infty.$

Answer 6

$$\lim\limits_{n\to\infty}\frac{(\ln n)^k}{\sqrt{n}}$$ Let $t=\frac{1}{\sqrt n}$, then $$\lim\limits_{t\to 0} t\left(\ln\left(\frac{1}{t^2}\right)\right)^k$$ $$=\lim\limits_{t\to 0} t\left(\ln 1-\ln t^2\right)^k$$ $$=\lim\limits_{t\to 0} t\left(-2\ln t\right)^k$$ $$=2^k\lim\limits_{t\to 0} t\left(-\ln t\right)^k$$ Let $u=-\ln t$, then $$2^k\lim\limits_{u\to\infty} e^{-u} u^k$$ $$=2^k\lim\limits_{u\to\infty} \frac{u^k}{e^u}=0$$ Since $e^u$ grows much faster than $u^k$, it's clear that this limit is zero.