Supremum of $\frac{m}{(n^2 + m)} $

Question

Can someone please help me prove that the supremum of this fraction is 1? m,n are naturals. I have been trying to prove it by contradiction, but nothing significant came out of it. I succeeded proving it for the infimum=zero by contradiction though.

Answer 1

Notice that as we increase $n$, the function $f(n) = \frac{m}{m + n^2}$ is decreasing for any fixed $m$. Therefore, we care about the set $A$ defined by $\left\{ \frac{m}{m + 1} : m \in \mathbb{N} \right\}$. Clearly the set $A$ is nonempty and bounded from above, since $m \ge x$ for any $x \in A$. Note that the original set is also bounded by $m$ and nonempty, so it doesn't matter if we focus on a subset.

To prove that $1$ is the supremum of $A$, we have to do two things:

1. We have to show that $1$ is an upper bound of $A$, and

2. We have to show that if $a$ is an upper bound of $A$, $a \ge 1$.

Note that $m + 1 \ge m$, so $\frac{m + 1}{m + 1} = 1 \ge \frac{m}{m + 1}$, and so $1$ is an upper bound of $A$. Now suppose $a$ is an upper bound of $A$ as well, but assume, to the contrary, that $1 > a$. Then $$1 > a \ge \frac{m}{m + 1},$$ for any $m \in \mathbb{N}$. Now note that $\displaystyle\lim_{m \to \infty} \frac{m}{m+1} = 1$. That is, for every $\varepsilon > 0$, there is some $N > 0$ such that $\left| \frac{m}{m + 1} - 1 \right| < \varepsilon$ whenever $m > N$. In other words, it must be true for the positive number $\varepsilon = 1 - a$, so we have $$-(1 - a) + 1 = a < \frac{m}{m + 1} < (1 - a) + 1 = 2 - a.$$ But we just said that $a \ge \frac{m}{m + 1}$, so we have reached a contradiction, and it cannot be the case that $1 - a > 0$, and so $1 - a \le 0$ or $1 \le a$.