Calculate sup and inf of $A=\{n/(1-n^2)\mid n>1\}.$

Question

Calculate $\sup$ and $\inf$ of the set, $$A=\{n/(1-n^2)\mid n>1\}.$$ I know that $\sup A=0$ and the $\inf A$ don't exist. I prove that $n/(1-n^2)<0$, but I can't prove that $0-\epsilon<n/(1-n^2)$, any help?

Answer 1

If we take the derivative $$\big(\frac{x}{1-x^2}\big)' = \frac{1+x^2}{(1-x^2)^2}$$, we see that the derivative is always positive for $x > 1$, so the sequence $\frac {n}{1-n^2}$ is increasing. Therefore its infimum is the first value, $-\frac{2}{3}$.

Answer 2

For $ x>1$,put $$f(x)=\frac{x}{1-x^2}$$ $$f'(x)=\frac{1+x^2}{(1-x^2)^2}>0$$ So $$\inf A =f(2)=-\frac 23$$ and if $ u_n= f(n)$ then

$$\sup A=\lim_{n\to+\infty}u_n=0$$

I used this:

If a sequence of reals $(u_n)_{n\ge n_0}$ is increasing and converges to $ L $

Then

$$\sup\{u_n\;:\; n\ge n_0\}=L$$

Answer 3

Define $f(n) = \dfrac{-n}{n^2 - 1}$, if $a < b$ so $f(a) < f(b)$: $$ab(a - b) < 0 < b - a$$ $$-ab^2 + a < -ba^2 + b$$ $$-a(b^2 - 1) < -b(a^2 - 1)$$ $$\dfrac{-a}{a^2 - 1} < \dfrac{-b}{b^2-1}$$ $$f(a) < f(b)$$

So $inf(A) = f(2) = -\dfrac{2}{3}$.

Now suppose $sup(A) = d < 0$, we know that $\lim_{n \rightarrow} f(n) = 0$, for all $\epsilon >0$ there exists $n_0$ such that $|f(n)| < \epsilon$ if $n > n_0$, choose $\epsilon = d/2$, so $|f(n)| < d/2 \longrightarrow f(n_0 + 1) > d/2 > d$.