Find the supremum and infimum of a set X

Question

Let the set $X=\{\frac{s}{s^2+2}, s\in\mathbb{Z}\}$ I have to find the supremum and the infimum. I have done in this way:
I can observe that $X=\{\frac{s}{s^2+2}, s\in\mathbb{N}\}\cup \{\frac{-t}{t^2+2}, t\in\mathbb{N}\}$ and now I study separetely the infimum and supremum of the two sets that give me in the union the set X.
1)$\{x_s=\frac{s}{s^2+2}, s\in\mathbb{N}\}$: since $\frac{s}{s^2+2}>\frac{s+1}{(s+1)^2+2}$ $\forall s\geq 1$ then my sequence is eventually decreasing and so $\exists \lim_{s\to\infty} x_s=0=inf$. The sup is instead given by $\frac{s}{s^2+2}$ evaluated at s=1, so $sup=\frac{1}{3}$.
2) $ \{x_t=\frac{-t}{t^2+2}, s\in\mathbb{N}\}$:
since $\frac{-t}{t^2+2}>-\frac{t+1}{(t+1)^2+2}$ $\forall t\geq 1$ then my sequence is eventually increasing and so $\exists \lim_{t\to\infty} x_t=0=sup$. The inf is instead given by $\frac{-t}{t^2+2}$ evaluated at t=1, so $inf=\frac{-1}{3}$.
So finally $supA=max\{\frac{1}{3},0\}=\frac{1}{3}$ and then $infA=max\{\frac{-1}{3},0\}=\frac{-1}{3}$.
TO DO: check my solving.

Answer 1

hint

Let $$f(x)=\frac{x}{x^2+2}$$

$ f $ is an odd function.

$$(\forall x\ge 0)\; \;\; f'(x)(x^2+2)^2=2-x^2$$

From here the maximum of $ X $ will be $ \max X =f(1) = f(2) =\frac 13$ .

By symmetry, $\min X =-\max X $.

Answer 2

Let $f(s) = s/(s^2 + 2)$. Then clearly for all $s \in \mathbb Z$, $$f(-s) = -s/((-s)^2 + 2) = -s/(s^2 + 2) = -f(s).$$ So if $f(n) = \sup X$, we must have $f(-n) = \inf X$.

Next, observe that $f(s) > 0$ if $s > 0$, so $f(s) < 0$ if $s < 0$, and $f(s) = 0$ if $s = 0$. So it suffices to consider $s \in \mathbb Z^+$ and $\sup X$ only.

Finally, consider the quotient $$\frac{f(s+1)}{f(s)} = \frac{(s+1)(s^2 + 2)}{s(s^2 + 2s + 3)} = \frac{s^3 + s^2 + 2s + 2}{s^3 + 2s^2 + 3s} = 1 - \frac{(s-1)(s+2)}{s(s^2 + 2s + 3)}.$$ When $s > 0$, the denominator of the second term is always positive. Since the numerator of the second term is zero if $s = 1$ (we ignore the negative root since $s > 0$), and is positive for $s > 1$, it follows that this ratio is equal to $1$ only when $s = 1$, and is strictly less than $1$ if $s > 1$. Therefore, for $s \in \mathbb Z^+$, we see $f(s)$ is greatest when $s = 1$, and $\sup X = f(1) = \frac{1}{3}$, from which it follows that $\inf X = -\frac{1}{3}$.