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.

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 $0$.  
