How can I prove $\lim_{n \to \infty} \frac{n^2}{2^n}=0$ How can I prove $\lim_{n \to \infty} \frac{n^2}{2^n}=0$
I tried to use $\mid \frac{n^2}{2^n} - 0 \mid <\epsilon $ However, because of $n^2$ I cannot use it. Also, I tried to use ratio test and I got $\lim_{n->\infty}\frac{(n+1)^2}{2}\frac{1}{n^2}$, but after that one, I dont know how to get limit < 0
I have no idea to solve this problem. 
I can also use a squeez lemma, but dont know how to apply to this question. 


Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:

$$
\lim_{n \to \infty}{n \over 2^{n}} =
\lim_{n \to \infty}{\left(n + 1\right) - n \over 2^{n + 1} - 2^{n}} =
\lim_{n \to \infty}{1 \over 2^{n}} = \bbox[#ffe,10px,border:1px dotted navy]{\displaystyle{\large 0}}
$$
