How to prove that $\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0$ & $\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0$?

Question

Prove that:

$$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$

I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?

What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$

What to do with $n!$?

Answer 1

Fact: If $a_n > 0 $ and $\lim_{n\to\infty}\Big|\frac{a_{n+1}}{a_n}\Big| = c < 1$ then $\lim_{n\to\infty}a_n = 0$.

Let $a_n = \frac{n^2}{2^n}$

$$\lim_{n\to\infty}\Big|\frac{a_{n+1}}{a_n}\Big|=\lim_{n\to\infty} \Big|\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}\Big| = \lim \frac{1+\frac{2}{n}+\frac{1}{n^2}}{2} = \frac{1}{2} < 1$$

If $\lim_{n\to\infty}\Big|\frac{a_{n+1}}{a_n}\Big| = c < 1$ then $\lim_{n\to\infty}a_n = 0$.

The second follows using the exact same idea.

Answer 2

Hint: Use L'Hopital's rule. Recall that the derivative of $2^n$ is $\ln 2 (2^n)$:

$$y=2^n$$ $$\ln y=n\ln2$$ $$\frac{y'}{y}=\ln2.$$

Answer 3

L'Hopital is too strong for this. Completely elementary, you can prove $2^n>n^3$ by induction: $2^n \times 2>n^3+3n^2+3n+1$ follows from $2^n>3n^2+3n+1$ which follows from $2^n>7n^2$ (for example). Using induction again, you can reduce this to $2^n>10n$ and then to $2^n>10$.