How to find the supremum and infimum of the following set of numbers

Question

Find the supremum and infimum of the set of numbers $\frac{(m+n)^2}{2^{mn}}$, where $m,n$ are natural numbers.

I know that lower bound is $0$. How to find the supremum?

Answer 1

Clearly, $0\leq (m+n)^2$ and $0\leq 2^{nm}$, thus $0\leq \frac{(n+m)^2}{2^{nm}}$ which means $0$ is a lower bound.

(Considering the situation when the set of natural number does not contain $0$)

Let $\epsilon>0$ be given and note that $\lim_{m\rightarrow \infty} \frac{(1+m)^2}{2^m}=0$ so there exists $M$ such that $\frac{(1+m)^2}{2^{m}}<\epsilon=0+\epsilon$ for any $m\geq M$. It implies $0$ is the infimum.

Let, focus on the diagonal line $y=-x+(n+m)$ which maintain the value of $x+y$ in the line. Then let $f(x,y)=\frac{(x+y)^2}{2^{xy}}$. We want to find the maximum value of $f(x,y)$ on the line where $(x,y)\in \mathbb{N}\times \mathbb{N}$. In the line, the maximum value occur when $2^{xy}$ is minimum which is when $(x,y)= (1,n+m-1)$ or $(n+m-1,1)$. Thus we get $$\frac{(n+m)^2}{2^{nm}}\leq f(1,n+m-1)=\frac{(n+m)^2}{2^{n+m-1}} \hspace{1cm}\forall (n,m)\in \mathbb{N}\times \mathbb{N}$$

Note that, by simple calculus, we conclude that the maximum value of of $f(x)=\frac{x^2}{2^{x-1}}$ occurs when $x=\frac{2}{\ln 2}$ which is close either $2$, $4$ or $3$. When $x=2$ or $4$, $f(x)=2$ and $f(3)=\frac{9}{4}$. Thus, on $\mathbb{N}\times \mathbb{N}$, $\frac{9}{4}$ is the maximum and supremum value. ($f$ is continuous and $[1,4]\times[1,4]\cap \mathbb{N}\times \mathbb{N}$ is compact, $f$ has maximum )

Answer 2

Let us check some values first.

$$ f(m,n)= \frac{(m+n)^2}{2^{mn}}$$

$$f(1,1)=2$$

$$f(1,2)=f(2,1)=2.25$$ $$ f(2,2) = 1$$ $$f(1,3)=f(3,1) = 1$$

$$f(2,3)=f(3,2) = \frac {25}{64}$$ Note that the function is a decreasing function of $m$ and $n$ once you get after $$f(1,2) = 2.25$$ which is the maximum value hence the supremum.

We may check the partial derivatives of $f$ with respect to $m$ and $n$ to verify the above statement.

The infimum is $0$ because the limit as $$(m,n)\to (\infty, \infty)$$ is $0$