# Show that $\lim_{n->\infty} \frac{n^2}{2^n} = 0$

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1. Feb 28, 2017

### kwangiyu

1. The problem statement, all variables and given/known data
show that
$$\lim_{n->\infty} \frac{n^2}{2^n} = 0$$

2. Relevant equations

squeeze theorem

3. The attempt at a solution

I tried to use squeez theorem. I dont know how to do it because dont know how to reduce $2^n$

However, I can solve this question like this.

Given $\epsilon>0$, find $M \in N$ such that $M > max (4, \frac{1}{\epsilon})$ $\mid \frac{n^2}{2^n} \mid = \frac{n^2}{2^n} < \frac{n}{n^2} = \frac{1}{n} < \frac{1}{M} <\epsilon$

$2^n > n^2$ and $if x>4$

My question is how can I solve this problem with squeez theorem ?

Last edited by a moderator: Mar 1, 2017
2. Feb 28, 2017

### Staff: Mentor

What about $2^n > n^3$ ?
And I think you also need it in your prove. How became $n^2 < n$ in the above?

3. Feb 28, 2017

### kwangiyu

oh.. I made mistake. but 2^n < n^3 isn't it ?

4. Feb 28, 2017

### Staff: Mentor

Yes, although you then need a higher lower bound for $M$. $4$ won't do anymore, but this isn't a problem.
And with that, your proof works and you can also use the same estimations for the squeeze theorem.

5. Mar 1, 2017

### Staff: Mentor

Is there some reason you need to use the squeeze theorem?

The expression you're taking the limit of can be written as $\left(\frac{n^{2/n}} 2\right)^n$
The problem then boils down to showing that $\frac{n^{2/n}} 2 < 1$, which can be done using induction.