TITLE: limit question: $\lim\limits_{n\to \infty } \frac{n}{2^n}=0$
QUESTION [8 upvotes]: $$ \lim_{n\to\infty}\frac n{2^n}=0. $$
I know how to prove it by using the trick, $2^n=(1+1)^n=1+n+\frac{n(n-1)}{2}+\text{...}$
But how to prove it without using this?

REPLY [3 votes]: By AM/GM:
$$\frac{2^{n}-1}{n}=\frac{2^0+2^1+\ldots +2^{n-1}}{n}\geq \left(2^{0+1+\ldots+(n-1)}\right)^{\frac1n}=2^{\frac{n-1}{2}}$$
Therefore,
$$0<\frac{n}{2^n}<\frac{n}{2^n-1}\leq 2^{-\frac{n-1}{2}}\rightarrow 0\quad\text{as}\quad
n\rightarrow\infty.$$

REPLY [3 votes]: One more answer with another approach I've used several times in this site and I'm surprised nobody's yet used. Put
$$a_n:=\frac n{2^n}\implies \frac{a_{n+1}}{a_n}=\frac{n+1}{2^{n+1}}\frac{2^n}n=\frac12\frac{n+1}n\xrightarrow [n\to\infty]{}\frac12$$
and thus the quotient (D'Alembert's) test gives us that the positive series
$$\sum_{n=1}^\infty\frac n{2^n}\;\;\text{converges}\implies \lim_{n\to\infty}\frac n{2^n}=0$$