$\lim\limits_{n\to \infty}\frac{n^k}{2^n}$ without l'Hopital

Question

Let $\varepsilon >0$. Let $N>?$. For any integer $n>N$ we have $$\frac{n^k}{2^n}<\varepsilon.$$ I don't know how to proceed here sensically.

I'd say we have to start with

$$<\frac{n^k}{2^N}$$

But what do we do here about the $n^k$?

Remember, I don't want to use l'Hopital or for me unproven limit laws like "exponents grow faster than powers"

Also, I don't want to use hidden l'Hopital, i.e. argumenting with derivatives. Since we don't even have proven derivative laws.

Consider $\dfrac{a_{n+1}}{a_n}$, where $a_n = \dfrac{n^k}{2^n}$. — Daniel Fischer, Dec 03 '16 at 10:45
Duplicate: How to prove that exponential grows faster than polynomial? (Found using Approach0.xyz) — Workaholic, Dec 03 '16 at 14:49
@Workaholic thanks for the site-tip. Could you mark as duplicate? — SAJW, Dec 03 '16 at 14:56

score 4 · Accepted Answer · answered Dec 03 '16 at 10:46

4

Let's take the logarithm. One gets

$$\log{n^k\over 2^n}=k\log{n}-n\log{2}=-n\left(\log{2}-k{\log{n}\over n}\right)$$

Now when $n\to\infty$ one has $\log{n}/n\to 0$ and so

$$\lim_{n\to\infty}\log{n^k\over 2^n}=-\infty$$

And so

$$\lim_{n\to\infty}{n^k\over 2^n}=0$$

answered Dec 03 '16 at 10:46

marwalix

Can you explain why we can conclude the last one? what is the reverse of taking the logarithm? If we have log$(a)=-\infty$ then we have something like $a^{-\infty}=0$? – SAJW Dec 03 '16 at 11:08
We are just using $\lim_{-\infty}e^x=0$ and the inverse function of the logarithm is the exponential – marwalix Dec 03 '16 at 11:42
but the log is base 2 would that make any difference? and exponential needs ln right? – Chaitanya Bapat Jan 27 '19 at 21:41

1 Answers1