inf A of $n^22^{-n}$ Let $A=\{n^22^{-n}, n \in \mathbb{N}\}$. Find $\inf A$, $\sup A$.
I tried starting by proving that $\frac{n^2}{2^n} \leq 1/n$ by induction. After, I showed that $\frac{n^2}{2^n} \geq 0$. By the squeeze theorem,
$$0 \geq \frac{n^2}{2^n} \geq \frac{1}{n}.$$
But my solution is awfully complicated : I used strong induction with $n+5$... Is there an easier way?
 A: $$
\begin{array}{r|l}
n & n^2 2^{-n} \\
\hline
1 & \frac12 \\[6pt]
2 & 1 \\[6pt]
3 & 1 + \frac18 \\[6pt]
4 & 1 \\[6pt]
5 & \frac{25}{32} \\[6pt]
6 & {}\ \ \vdots
\end{array}
$$
If you can show that it keeps getting smaller after that, then you've got the supremum.
If you can show the limit is $0$ and every term is positive, then you've got the infimum.
If $n\ge 4$, then every time $n$ is incremented by $1$, the numerator is multiplied by $\left(\frac{n+1}{n}\right)^2=\left(1+\frac1n\right)^2 \le \frac{25}{16} = 1.5625$ and the denominator is multiplied by $2$.  That is certainly enough to show that the sequence decreases after that, so the sup is the max, $1+\frac18$.
At every step one multiplies by something $\le \frac{25}{32}$, and the problem is to show that the inf is $0$.  If, not then the inf is some positive number $c$, and $0<c\le\left(\frac{25}{32}\right)^n$ for all positive integers $n$.  So $1/c>\left(\frac{32}{25}\right)^n$ for all $n$.  Hence the following exists:
$$
d=\sup\left\{ \left(\frac{32}{25}\right)^n : n=1,2,3,\ldots \right\}.
$$
Since $d\cdot\frac{25}{32}<d$, then $d\cdot\frac{25}{32}$ must not be an upper bound of that set.  Consequently for some $n$ $d\cdot\frac{25}{32}\le \left(\frac{32}{25}\right)^n$.  But then $d\le\left(\frac{32}{25}\right)^{n+1}$ and we get a contradiction.
The assumption that the inf is $>0$ led to a contradiction.
