Prove that $A = \bigg \{ \frac{n}{1 - n^2} \hspace{.1cm} | \hspace{.1cm} n \in \mathbb{N}, n \ge 2 \bigg \}$ has a supremum equal to $0$.

Question

Consider the set:

$$A = \bigg \{ \frac{n}{1 - n^2} \hspace{.1cm} | \hspace{.1cm} n \in \mathbb{N}, n \ge 2 \bigg \}$$

I have to show that the supremum of this set is $0$, using the analysis introductory tools only (so I can't use limits, and things like that). This is what I tried.

We know

$$\frac{n}{1 - n^ 2} \le 0, \hspace{0.1cm} \forall n \in \mathbb{N}, \hspace{0.1cm} n \ge 2$$

$$=> 0 \in ub(A) => A \hspace{0.1cm} \text{bounded above (1)}$$

where $ub(A)$ is the set of upper bounds of $A$. Because we have condition $(1)$ and we know that $A \ne \emptyset$, we can conclude by the Supremum Property that

$$\exists \hspace{0.1cm} sup(A) \in \mathbb{R}, \hspace{0.1cm} sup(A) \le 0$$

where I got the condition that $sup(A) \le 0$ from the fact that $0$ is an upper bound.

So we now know that $sup(A) \le 0$. I want to show that it is equal to $0$.

So let's suppose that $sup(A) \lt 0, \hspace{0.1cm} sup(A) \in \mathbb{R}$

Since the supremum is an upper bound (the smallest upper bound), we know

$$sup(A) \ge \frac{n}{1 - n^2}, \hspace{0.1cm} \forall n\in \mathbb{N}, \hspace{0.1cm} n \ge 2$$

Through a simple calculation, we can also show

$$\frac{n}{1 - n^2} \gt - \frac{1}{n^2}, \hspace{0.1cm} \forall n \in \mathbb{N}, n \ge 2$$

Take $\varepsilon = \sqrt{- \frac{sup(A)}{2}}$

We can do this, since we supposed $sup(A) \lt 0$, so then $- \frac{sup(A)}{2} \gt 0$

$$\text{By the Archimedian Property => $\exists \hspace{0.1cm} n\in \mathbb{N}$ s.t. $n \gt \frac{1}{\varepsilon}$}$$

$$=> \exists \hspace{0.1cm} n \in \mathbb{N} \hspace{0.1cm} \text{s.t.} \hspace{0.1cm} n > \sqrt{\frac{-2}{sup(A)}}$$

Both sides are positive, so we can square the expression without a sign change.

$$n^2 \gt \frac{-2}{sup(A)}$$

$$\frac{1}{n^2} \lt \frac{-sup(A)}{2}$$

$$=> \exists \hspace{0.1cm} n \in \mathbb{N} \hspace{0.1cm} \text{s.t.} \hspace{0.1cm} - \frac{1}{n^2} \gt \frac{sup(A)}{2}$$

For $n = 1$, we'd have $sup(A) < -2$, which cannot be true since we can show that $min(A) = - \frac{2}{3}$. So $n \in \mathbb{N}$, $n \ge 2$. So now we are within the conditions of $A$ (regarding the number $n$).

Now we have:

$$sup(A) \ge \frac{n}{1 - n^2} \ge - \frac{1}{n^2} \gt \frac{sup(A)}{2}$$

So we have

$$sup(A) \gt \frac{sup(A)}{2}$$

But our assumption was that $sup(A) \lt 0$. So $sup(A)$ is negative and that makes the above inequality false. Thus, we have a contradiction, so our assumption is false. So we can conclude that $sup(A) = 0$.

Can this be considered correct? Or did I screw things up? Did I somehow unnecessarly complicate things? Should I have proved this differently?

Answer 1

Your solution looks fine to me, however I think it can be done in a shorter way.

It suffices to show that $\frac{n}{n^2-1}$ can be made arbitrarily small. WLOG suppose $n \ge 2$, so that $n - \frac{1}{n} \ge \frac{n}{2}$. We have $$ \frac{n}{n^2-1} = \frac{1}{n - \frac{1}{n}} \le \frac{1}{\frac{n}{2}} = \frac{2}{n}$$ and then apply archimedian.

Answer 2

For $n \ge 2$ $$ 0 <\frac{n}{n^2-1}=\frac{n}{n+1}\frac{1}{n-1} < \frac{1}{n-1} \\ $$ Multiply by $-1$ $$ 0 > \frac{n}{1-n^2} > \frac{1}{1-n} \\ $$