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

[kwangiyu]

Homework Statement

show that
\lim_{n->\infty} \frac{n^2}{2^n} = 0

Homework Equations

squeeze theorem

The Attempt at a Solution

I tried to use squeez theorem. I don't know how to do it because don't 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>4My question is how can I solve this problem with squeez theorem ?

 

[fresh_42]

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

 

[kwangiyu]

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

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

 

[fresh_42]

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.

 

[Mark44]

Homework Statement

show that
\lim_{n-&gt;\infty} \frac{n^2}{2^n} = 0

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

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.

 

[chwala]

Looks like the OP did not state/show whether he could proceed to solution...no follow up on his part.