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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.

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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 $.

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  • $\begingroup$ Ok thanks but I would like also to know if my solving is correct $\endgroup$ – Nik Jan 13 at 20:55
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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 $\inf X = -\frac{1}{3}$.

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  • $\begingroup$ Ok thanks but I would like also to know if my solving is correct $\endgroup$ – Nik Jan 13 at 20:49
  • $\begingroup$ @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., $$X = \left\{ \frac{s}{s^2 + 2} : s \in \mathbb N \right\} \cup \left\{\frac{-t}{t^2 + 2} : \color{red}{t} \in \mathbb N \right\}.$$ $\endgroup$ – heropup Jan 13 at 21:37
  • $\begingroup$ 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 $\forall s\geq 1$. $\endgroup$ – Nik Jan 13 at 21:44

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