Show that $\left\{\frac{m}{mn+1}: m,n\in\Bbb N*\right\}$ is bounded

Question

Let $$A = \left\{\frac{m}{mn+1}: m,n\in\Bbb N*\right\}$$

The question is as follows: Show that A has an upper bound and a lower bound and determine them.

My attempt:

I first thought of giving inequations for m and n since they are in $\Bbb N*$, so
$m>0$ and $n>0$
meaning $$mn>0$$ $$mn+1>1$$ $$0<\frac{1}{mn+1}<1$$ Now if we multiply each by the strictly positive integer $m$ we get $$0<\frac{m}{mn+1}<m$$ Now we can see that $\inf A$ exists (while i am not sure how am i going to get it) Besides that i can't tell that A has an upper bound since $m$ is not a constant, i tried some way rather than multiplying and no conclusion.
So help me please and tell me if my start is right, I appreciate your opinions

Answer 1

You write:

I first thought of giving inequations for m and n since they are in N∗, so $m>0$ and $n>0$ meaning $mn>0$ $mn+1>1$ $0<\frac 1{mn+1}<1$ Now if we multiply each by the strictly positive integer $m$ we get $0<\frac m{mn+1}<m$

You can do better by considering $n \ge 1$ so

$mn \ge m$ and

$mn + 1 > m$ and so $0 < \frac 1{mn+1} < \frac 1m$ and $0 < \frac m{mn+1} < 1$.

Now you can see that $A$ must be bounded below by $0$ and above by $1$.

To find the $\inf$ it might help to note that $\frac m{mn+1} = \frac 1{n+\frac 1m}$.

$\frac 1{n+\frac 1m} < \frac 1n$ so $\inf A \le \inf \{\frac 1n\}$.

As for the $\sup$ it might help to note that $B= \{\frac m{m\cdot 1 + 1}\}=\{\frac m{m+1}\}\subset A$ so $\sup A \ge \{\frac m{m+1}\}$.