How can I prove that sup(A)=infinity in this situation?

Question

If I have set $$A=\Big\{(n+m^2)/(n^2+m) \; : \; n,m\in\mathbb{N}\Big\},$$ and I would like to prove that $\sup(A)=\infty$.

Currently my approach is to set $n=1$ and assume towards contradiction that set $A$ is bound from above. I thought to use the Archimedean property to show that since $m,m^2\in\mathbb{N}$ and $m^2>m$, dividing them as the value of $m$ increases will result in values that are infinitely increasing and therefore the equation has no upper bound. Is this a good approach, and if not how might I approach this? Thanks in advance!

Answer 1

This is a good approach, you just need to finish it through. If you assumed $A$ was bounded by the positive real number $\alpha$ you'd have for every natural number $m$ the inequalities $$m^2 \leqslant 1+m^2 \leqslant (1+m)\alpha \leqslant (m+m)\alpha = 2\alpha m,$$ whereby we set $n=1$. Now you can divide by $m$ to find that $m \leqslant 2\alpha$ for every natural number $m$, which is clearly false.

Answer 2

Your approach is surely a good one.

Here is another approach:

Let $n=1$, then

$$\frac{n+m^2}{n^2+m} = \frac{1+m^2}{1+m} \gt \frac{m^2-1}{m+1} = m-1 $$

Since $m$ is an arbitrary natural number, $m-1$ is unbounded. $\therefore \frac{1+m^2}{1+m}$ is unbounded.

$\therefore \sup(A)=\infty$

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Note: One may also use

$$\lim_{m \rightarrow \infty}\frac{1+m^2}{1+m}=\lim_{m \rightarrow \infty}\frac{\frac{1}{m^2}+1}{\frac{1}{m^2}+\frac{1}{m}}=\infty$$

to claim that $\frac{1+m^2}{1+m}$ is unbounded.