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

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?

• Obviously zero is a lower bound. Did you prove that it is the infimum? Jun 17, 2019 at 18:09
• Lower bound is $0$ if $m,n \in \mathbb{N}$, i.e. the lower bound and infimum are not the same thing. Jun 17, 2019 at 18:09
• @Coltrane Because for every $m,n$ the quantity is greater than zero, so zero is a lower bound. Jun 17, 2019 at 18:09
• Is zero considered natural? In this case the set has infinity as supremum. Jun 17, 2019 at 18:10
• You're right that's what I meant Jun 17, 2019 at 18:10

## 2 Answers

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 )

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 $$0$$