Finding $\inf(S)$ and $\sup(S)$ for $S = \{\frac{(m+n)^2}{2^{mn}} : m,n \in \mathbb{N}\}$

Question

Given a set S such that $S = \{\frac{(m+n)^2}{2^{mn}} : m,n \in \mathbb{N}\}$, what is $\sup{S}$ and what is $\inf{S}$?

I think $\inf{S}$ is 0 because if we let $n = m$ and allow $m$ to approach infinity, the limit is 0, and clearly no element of $S$ can be negative. However, I am not sure how to find the supremum, except

Answer 1

Notice that if you increase $n$ (or equivalently $m$) by $1$, we have $$\frac{\frac{(m+n+1)^{2}}{2^{m(n+1)}}}{\frac{(m+n)^{2}}{2^{mn}}}=\left(1+\frac{1}{m+n}\right)^{2}\frac{1}{2^{m}}$$ If this is less than $1$ (and it is if $m$ or $n$ is at least $2$), then we lose by making $m$ or $n$ any bigger. This leaves a limited number of cases to check.

Answer 2

Except for $(2,2)$ the $S$ is less than $1$ that's the reason why limit approaches to $0$ after equalizing $m,n$.So sup(S)=1