Limit of $\frac{n^2}{4^n}$

Question

How can I prove that the limit of $n^2/4^n$ as $n$ approaches infinity is 0? I want to solve it without using de l'Hospital's rule and I tried some inequalities, but I don't find a nice solution.

Answer 1

A simple way $$ a_n=\frac{n^2}{4^n}=\left(\frac{\sqrt{n}}{(\sqrt{2})^n}\right)^4. $$ But $$ (\sqrt{2})^n=\big(1+(\sqrt{2}-1)\big)^n\ge 1+n(\sqrt{2}-1), $$ and hence $$ 0<a_n\le \left(\frac{\sqrt{n}}{1+n(\sqrt{2}-1)}\right)^4\to 0. $$

We have used the inequality $$ (1+a)^n\ge 1+na, $$ whenever $a>-1$ and $n$ positive integer, which can be shown inductively.

Answer 2

Since $$\lim_{n\to \infty} \sqrt[n]{\frac{n^2}{4^n}} = \lim_{n\to \infty} \frac{n^{2/n}}{4} = \frac{1}{4} < 1,$$

by the ratio test, the series $\sum_{n = 1}^\infty \frac{n^2}{4^n}$ converges. So by the $n$th term test, $\lim_{n\to \infty} \frac{n^2}{4^n} = 0.$

Alternatively, consider the sequence $a_n = \frac{n}{2^n}$. By the binomial theorem, $2^{n+1} = (1 + 1)^{n+1} > \frac{(n+1)n}{2} > \frac{n^2}{2}$. So $$0 < a_n = 2\cdot \frac{n}{2^{n+1}} < 2\cdot \frac{2}{n} = \frac{4}{n}$$

Since $\lim\limits_{n\to \infty}\frac{4}{n} = 0$, it follows that $\lim\limits_{n\to \infty} a_n = 0$. Thus $$\lim_{n\to \infty} \frac{n^2}{4^n} = \lim_{n\to \infty} a_n^2 = \left(\lim_{n\to \infty} a_n\right)^2 = 0.$$

Answer 3

Let $x_n=\frac{n^2}{4^n}$

$\frac{x_{n+1}}{x_n}=\frac{(n+1)^2}{n^2}\frac{1}{4}$

Now, we know that $\lim_{n\to+\infty}\frac{(n+1)^2}{n^2}=1$

That is, for some $N$, $\frac{(n+1)^2}{n^2}<4$ for all $n>N$

Thus, $x_n$ is ultimately decreasing. We can assume it is always decreasing without loss of generality.

Now, denote the limit by $x$

$\lim_{n\to+\infty}x_n=\lim_{n\to+\infty}x_{n+1}$

$x=\frac{1}{4}x$

Thus $x=0$. QED.