# Proof for a Sequence Convergence

1. Dec 21, 2012

### Bachelier

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

2. Dec 21, 2012

### LCKurtz

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

3. Dec 21, 2012

### Bachelier

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.