Prove using the basic properties of a sequence that $\lim \frac{n}{2^n} = 0$ Prove using the basic properties of a sequence that $\lim \frac{n}{2^n} = 0$

I tried to prove this using the standard epsilon-definition of a limit but the math got really hard, so I stopped. Then, I tried to represent the sequence as a quotient or a product of two sequences, but those sequences were not convergent. Thus, I've got no more ideas on how to tackle this problem. Any suggestions would be most appreciated.
 A: By the binomial theorem $2^n = (1 + 1)^n > n(n-1)/2$ for all $n \ge 1$, so $$\frac{n}{2^n} < \frac{2}{n-1}$$
for all $n \ge 1$. Given $\epsilon > 0$, choose a positive integer $N > \dfrac{2}{\epsilon} + 1$, then use the above inequality to show that $\dfrac{n}{2^n} < \epsilon$ for all $n \ge N$.
A: As the comment suggests, you can notice that $2^{n}$ grows exponentially, and thus, eventually, we would have $2^{n}>n^{2}$, and thus the limit becomes sandwiched between $0$ and $\lim \frac{1}{n}$, and hence it becomes $0$.
Otherwise, you could also note that $2^{n}=e^{nlog(2)}=1+nlog(2)+\frac{1}{2!}(nlog(2))^{2}+..$, and so $\frac{n}{2^{n}}=\frac{1}{\frac{1}{n}+log(2)+n(..)}$, and when $n\rightarrow \infty$, the terms with a factor of $n$ in the denominator dominate and so the limit becomes $0$.
A: Hint:
$\dfrac{u_{n+1}}{u_n}\to \dfrac12$, hence $(u_n)$ is bounded from above by a geometric sequence of ratio $<1$ is $n$ is large enough (actually you should find a geometric series of ratio $\dfrac34$ if $n>1$).
