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. 
 A: Notice that this can be seen to be equivalent to proving that the following converges:
$$\sum_{k=1}^\infty a_k,\quad a_n=\frac{n^2}{2^n}$$
By the ratio test, we need to show that
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$
but we can see that
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2n^2}$$
which is where you were at.  Now notice that
$$\frac{(n+1)^2}{2n^2}=\frac12\left(1+\frac1n\right)^2\stackrel{n\to\infty}\longrightarrow\frac12<1$$
A: One easy way is to use the binomial theorem to write for $n\ge 3$
$$2^n=(1+1)^n=\sum_{k=1}^n\binom{n}{k}\ge \frac{1}{6}n(n-1)(n-2)$$
Then we have
$$\frac{n^2}{2^n}\le \frac{6}{(n-1)(1-2/n)}$$
Can you finish now?

Another way is to note that $2^n=e^{n\log(2)}\ge \frac16 (\log(2)n)^3$ so that 
$$\frac{n^2}{2^n}\le \frac{6}{\log^3(2) n}$$

A third way is to observe that 
$$\lim_{n\to \infty}\left(\frac{n^2}{2^n}\right)^{1/n}=\frac12$$
Thus, for any $\epsilon>0$ there exists a number $N$ such that 
$$\left(\frac{n^2}{2^n}\right)\le \left(\frac{1}{2}+\epsilon\right)^n$$
whenever $n>N$.  Take $\epsilon=1/4$.  Then, there exists a number $N$ such that 
$$\left(\frac{n^2}{2^n}\right)\le \left(\frac34\right)^n$$
whenever $n>N$.
A: If $a_n=\frac{n^2}{2^n}$ then 
$$ \frac{a_{n+1}}{a_n}=\frac{(1+1/n)^2}{2}$$
For $n>2$, show $$\frac{(1+1/n)^2}{2}\leq \frac{8}{9}$$
So $$a_{n+3}\leq \left(\frac{8}{9}\right)^n a_3$$
A: We know $$\ell=\lim_{n \to \infty} \dfrac{n}{2^n}=\lim_{n \to \infty} \dfrac{n+1}{2^{n+1}}=\frac12\lim_{n \to \infty} \dfrac{n}{2^n}+\lim_{n \to \infty} \dfrac{1}{2^{n+1}}=\frac12\ell+0=\frac12\ell$$
so $\ell=0$. 
$$\ell=\lim_{n \to \infty} \frac{n^2}{2^n}=\lim_{n \to \infty} \frac{(n+1)^2}{2^{n+1}}=\frac12\lim_{n \to \infty} \frac{n^2}{2^n}+\lim_{n \to \infty} \frac{n}{2^n}+\lim_{n \to \infty} \frac{1}{2^{n+1}}=\frac12\ell+0+0$$
so $\ell=0$. 
A: 
Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:

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