Find the infimum and supremum(if they exist) of the set:
$$S=\{\frac{mn}{1+m+n}: m,n\in\Bbb N\}$$
I know
$$\frac{mn}{1+m+n}>0, \forall m,n\in\Bbb N$$
so $0$ is a lower bound. Now I have to find some subsequence whose limit is $0$ to prove $0$ is $\inf S$.
Let $n=\frac{1}{m}$. Then $$\lim_{m\to\infty}\frac{mn}{1+m+n}=\lim_{m\to\infty}\frac{1}{1+m+\frac{1}{m}}=0$$
So $\inf S=0$.
Is this correct? And how do I now find $\sup S$?
Best Answer
$$x=\frac{mn}{1+m+n}\implies x+1=\frac{(m+1)(n+1)}{1+m+n}\implies\frac{m+1}{m}\frac{n+1}{n}=\frac{x+1}{x}$$
Now, note that for positive integer $k$, we have $1<\dfrac{k+1}{k}\le2$.
We can thus infer that $\dfrac{x+1}{x}$ takes values between $1$ and $4$, i.e.
$$1<\frac{x+1}{x}\le4 \implies 0<\frac{1}{x}\le3\implies x\in \left[\frac{1}{3},\infty\right)$$
We then see that the infimum is $\frac{1}{3}$ and the supremum is $\infty$.