calculus - 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.

Answer

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 $0$.