Find the supremum and the infimum of the following set o real numbers 
Let $A = \left\{ \dfrac{(n+m)^2}{2^{nm}} : n,m \in \mathbb{N} \right\}
 $. Find $\sup A $ and $\inf A$

Attempt:
My idea is to write $A_m = \left\{ \dfrac{(n+m)^2}{2^{nm}} : n\in \mathbb{N} \right\} $ and so $A = \bigcup_{m \geq 1} A_m $ So to find the sup, we know that $\sup A = \max_{m} (\sup A_m) $
So, we see that for $A_1 $, $\sup A_1 = 9/4$, and for $A_2$, $\sup A_2 = 9/4$, and $\sup A_3 = 2$ and eventually $\sup A_k = \dfrac{(n+k)^2 }{(2^k)^n } $ so clearly as $k$ gets bigger, $\sup A_k$ get smaller. Thus, $\max_m ( \sup A_m ) = \dfrac{9}{4} $ which implies that $\boxed{ \sup A = \dfrac{9}{4} }$
Finally, as for the infimum, we can observe that $\dfrac{(n+m)^2 }{2^{nm}} \geq 0$ for any $n,m$ and thus $\boxed{ \inf A = 0}$
Is my work correct?
 A: The basic idea for your first part will work, but as was noted in the comments your statement that $\sup A_k=\frac{(n+k)^2}{(2^k)^n}$ really doesn’t make sense: $n$ is completely undefined here. You also need to fill in more details to justify some of your steps. After you’ve defined
$$A_m=\left\{\dfrac{(n+m)^2}{2^{nm}}:n\in\Bbb Z^+\right\}\;,$$
let $f_m(x)=\frac{(x+m)^2}{2^{mx}}$; then 
$$\begin{align*}
f_m'(x)&=2(x+m)2^{-mx}-(x+m)^22^{-mx}\ln2\\
&=2^{-mx}(x+m)\big(2-(x+m)\ln2\big)\;,
\end{align*}$$
so $f_m(x)$ is increasing for for $x<\frac2{\ln 2}-m$ and decreasing for $x>\frac2{\ln 2}-m$. In particular, $\max A_1=f_1(2)=\frac94$, and $\max A_m=f_m(1)=\frac{(m+1)^2}{2^m}$ for $m>1$. A similar analysis shows that this attains its maximum at $m=2$, and $f_2(1)=\frac94$, so 
$$\max\{\max A_m:m\in\Bbb Z^+\}=\frac94\;.$$
In other words, $\sup A=\frac94$ and is attained both at $m=1,n=2$ and at $m=2,n=1$, so that we can legitmately write $\max A=\frac94$.
As Paul Sinclair noted in the comments, you have only half of the argument needed to show that $\inf A=0$. Specifically, you’ve shown that $0$ is a lower bound for $A$, so clearly $0\le\inf A$. To finish the job you need to show that for each $\epsilon>0$ there are $m,n\in\Bbb Z^+$ such that $\frac{(n+m)^2}{2^{nm}}<\epsilon$. How fussily careful you have to be depends on how what has already been proved and can be taken for granted; I suspect that in your context it would suffice to take $m=1$ and note that $\lim_\limits{n\to\infty}\frac{(n+1)^2}{2^n}=0$.
Finally, in general it really is better to use $\sup$ when there is any possible question whether the supremum is attained, and I could well have used it throughout the first part of my answer. In this case I thought it sufficiently clear that I was producing actual maxima, so I went ahead and used $\max$.
