Calculating $\lim\limits_{n\to\infty}\frac{(\ln n)^{2}}{n}$ What is the value of $\lim\limits_{n\to\infty}\dfrac{(\ln n)^{2}}{n}$ and the proof ?
I can't find anything related to it from questions.
Just only $\lim\limits_{n\to\infty}\dfrac{\ln n}{n}=0$, which I know it is proved by Cesàro.
 A: Let 


*

*$S = \displaystyle\lim_{n \to \infty} \frac{\log(n)}{\sqrt{n}}$. What do you think the value of this limit would be? What you want to find is $S^{2}$.

*Secondly to get bounds for this limit consider the function $f(x)= \frac{\log(x)}{\sqrt{x}}$ on the interval $[1,\infty)$ and apply the derivative tests. 
A: Using the power series expansion for $e^{x}$, we have that
$$n=e^{\ln n}=\sum_{k=0}^{\infty}\frac{(\ln n)^{k}}{k!}>\frac{(\ln n)^{k}}{k!},\forall k\in\mathbb{Z}^{\geq 0}$$
for all positive integers $n$. So for any positive integer $k$, 
$$\underbrace{(k!)^{\frac{1}{k}}}_{C_{k}}n^{\frac{1}{k}}>\ln n$$
If we take $k=3$, then 
$$\lim_{n\rightarrow\infty}\frac{(\ln n)^{2}}{n}\leq \lim_{n\rightarrow\infty}\frac{C_{3}^{2}n^{\frac{2}{3}}}{n}=C_{3}^{2}\lim_{n\rightarrow \infty}\frac{1}{n^{\frac{1}{3}}}=0$$
A: We can get L'Hospital's Rule to work in one step. Express $\dfrac{\log^2 x}{x}$ as $\dfrac{\log x}{\sqrt{x}}\cdot\dfrac{\log x}{\sqrt{x}}$. L'Hospital's Rule gives limit $0$ for each part.
Another approach is to let $x=e^y$. Then we want to find $\displaystyle\lim{y\to\infty} \dfrac{y^2}{e^y}$. 
Note that for positive $y$, we have 
$$e^y\gt 1+y+\frac{y^2}{2!}+\frac{y^3}{3!}\gt \frac{y^3}{3!}.$$ 
It follows that $\dfrac{y^2}{e^y}\lt \dfrac{3!}{y}$, which is enough to show that the limit is $0$.  
A: There's for sure an easier way, but if we use L'Hospital we get
\begin{align}
\frac{d}{dn}\Bigl[ \frac{\ln(n)^2}{n} \Bigr] &= -\frac{(\ln(n)-2)\ln(n)}{n^2}
\\ &=- \frac{\ln(n)^2}{n^2} + 2\frac{\ln(n)}{n^2}
\end{align}
And you should be able to continue from this, knowing something about $\frac{\ln(n)}{n}$.
