Find the supreme and infimum of the set $$A=\left \{\dfrac{3n}{\sqrt{1+2n^2}}: n\in \Bbb N\right \}$$
We claim that, $\inf A=\sqrt{3}$, y $\sup A=\dfrac{3\sqrt{2}}{2}=\dfrac{3}{\sqrt{2}}.$ Let's prove it.
Supremo: Let's first show that $\sup A=\dfrac{3\sqrt{2}}{2}.$. Indeed, we must show the following two conditions,
$a)$ $\dfrac{3\sqrt{2}}{2}$ is upper bound for $A$.
$b)$ For all $\epsilon>0$, exists $a\in A$ such that $$\dfrac{3\sqrt{2}}{2}-\epsilon<a. $$
To show $ a) $, let's notice the following, $$\begin{align*} 0&\leq 2\\ 4n^2&\leq 2+4n^2\\ 4n^2&\leq 2(1+2n^2)\\ 2n&\leq \sqrt{2}\sqrt{1+2n^2}\\ 6n&\leq 3\sqrt{2}\sqrt{1+2n^2}\\ \dfrac{3n}{\sqrt{1+2n^2}}&\leq \dfrac{3\sqrt{2}}{2}. \end{align*}$$ In this way, we have shown that $ a) $ is true.
Let's show $ b) $. Since $ n \in \mathbb N $, then clearly we have that $ 9n ^ 2> 1 + 2n ^ 2 $, from which it follows that $ 3n> \sqrt {1 + 2n ^ 2} $. Therefore, $$\begin{align*} \dfrac{2\sqrt{2}3n}{\sqrt{1+2n^2}}&>3\\ \dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}+\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}&>3\\ \dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}&>-\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}+3. \end{align*}$$
Now, for Archimedean property given $ \epsilon> 0 $ (in this case $ \sqrt {2} \epsilon> 0 $), there exists $ n \in \Bbb N $ such that $$\sqrt{2}\epsilon>\dfrac{1}{n}>\dfrac{1}{\sqrt{1+2n^2}}>\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}>-\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}+3.$$ Then, $$\begin{align*} -\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}+3&<\sqrt{2}\epsilon\\ -\dfrac{\sqrt{2}3n}{\sqrt{1+2n^2}}&<\sqrt{2}\epsilon-3\\ -\sqrt{2}3n&<\sqrt{2}\epsilon \sqrt{1+2n^2}-3\sqrt{1+2n^2}\\ 3\sqrt{1+2n^2}-\sqrt{2}3n&<\sqrt{2}\epsilon\sqrt{1+2n^2}\\ \dfrac{3}{\sqrt{2}}-\dfrac{3n}{\sqrt{1+2n^2}}&<\epsilon\\ \dfrac{3}{\sqrt{2}}-\epsilon&<\dfrac{3n}{\sqrt{1+2n^2}} \end{align*}$$ Which shows that $ \sup A = \dfrac {3 \sqrt {2}} {2} = \dfrac {3} {\sqrt {2}} $.
Is my proof correct? Any suggestions to make it simpler? And some suggestion also for the $ \epsilon $ but with the lowest.