Proving $\lim_{n \to \infty} \frac{n}{4^n}=0$

Question

Why is

$$\lim_{n \to \infty} \frac{n}{4^n}=0\text{ ?}$$

Having a hard time proving this one with the definitions of the limit of a sequence. I'm trying to apply Bernoulli's inequality to $2^n$ to make it easier for myself but I'm completely lost.

Answer 1

There is some $N \geq 1$ such that for all $n \geq N$ we have $$ \frac{n}{4^{n}} = \frac{n}{2^{2n}} < \frac{n}{2^{n}} \leq \frac{n}{n^{2}} = \frac{1}{n}; $$ if $\varepsilon > 0$, then $n \geq \lceil \frac{1}{\varepsilon} \rceil + 1$ only if $\frac{1}{n} < \varepsilon$; therefore, if $n \geq \max \{ N, \lceil \frac{1}{\varepsilon} \rceil + 1 \}$, then $\frac{1}{n} < \varepsilon$.

Answer 2

By the Bernoulli Inequality, we have $2^n\ge 1+n$ for every positive integer $n$, and therefore $4^n=2^n\cdot 2^n\ge (1+n)^2$. Thus if $n\ge 1$, then $$\frac{n}{4^n}\le \frac{n}{(n+1)^2}\lt \frac{n}{n^2}=\frac{1}{n}.$$ Now producing, for given $\epsilon\gt 0$, an appropriate $N$ should not be difficult.