# Find the supremum and infimum of a set X

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.

## 2 Answers

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

• Ok thanks but I would like also to know if my solving is correct – Nik Jan 13 at 20:55

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 $$\inf X = -\frac{1}{3}$$.

• Ok thanks but I would like also to know if my solving is correct – Nik Jan 13 at 20:49
• @Nik Your proof is basically the same reasoning as mine, except you are missing certain steps, which I have filled in, such as justifying why $f$ is decreasing for $s \ge 1$. And there are a few typos; e.g., $$X = \left\{ \frac{s}{s^2 + 2} : s \in \mathbb N \right\} \cup \left\{\frac{-t}{t^2 + 2} : \color{red}{t} \in \mathbb N \right\}.$$ – heropup Jan 13 at 21:37
• Thanks for the correction of $t$...about the decrasing I have solved the inequality $f(s+1)<f(s)$ that it is true for $s>\frac{1+\sqrt{5}}{2}$, so surely $\forall s\geq 1$. – Nik Jan 13 at 21:44