I am considering how to find a supremum and infimum of 
$\frac{(n+1)^{2}}{2^{n}}, n\in \mathbb{N}$. I proved with induction that for $n\geq 6$ it goes to zero so infimum is also zero. For other n $(=1, 2, 3, 4, 5)$ I checked that supremum is $\frac{9}{4}$. And now I would like to prove it with Archimedes rule using $\varepsilon $, but I end with $\frac{9}{4}- \varepsilon < \frac{(n+1)^{2}}{2^{n}}$, and I have no idea how to transform it to use Archimedes; something like $n> ...$ Any idea? Thanks in advance!