Proof for a Sequence Convergence

[Bachelier]

$\text{We need to prove that the sequence} \ a_{n} = \{n^{2}/2^{n}\} \ \text{converges to 0} \\<br /> <br /> \text{Consider the sequence {n/ 2n} = { 0, 1/2, 1/2, 3/8, 1/4, 5/32, ...}. The terms get smaller and smaller.}\\<br /> \\<br /> \text{we can easily show that} \ n/2^{n}<=1/n \ \forall n>3 \\<br /> \text{from the fact that} \ n^{2}<=2^{n} \ \text{(insert proof by induction here)}\\<br /> <br /> then \ for \ \epsilon > 0 \ choose \ N = max\{3,1/\epsilon\} \\<br /> <br /> \text{The idea is that for all n>N, we will have} \ 1/n<1/N< \epsilon. \\<br /> \text{The problem is I think N needs to be greater than the max. i.e. } \ N> max\{3,1/\epsilon\} \ for \ this \ to \ work \\<br /> \\<br /> \text{otherwise we'll end up with a case where} \ 1/N= \epsilon$

 

[LCKurtz]

So use $\epsilon/2$ to begin with or let $N = \hbox{max}\{3,1/\epsilon\}+1$.

 

[Bachelier]

So use $\epsilon/2$ to begin with or let $N = \hbox{max}\{3,1/\epsilon\}+1$.

OK. So my question is correct, putting $N = \hbox{max}\{3,1/\epsilon\}$ was incorrect.
I found this argument on a website and wasn't sure.
Thanks.