Verify the inf of a subset S of $\mathbb R$

Question

Let S=$\{a_n=\frac{3n+1}{n}:n\geq 0\}$ , verify that $infS=3$.
MY IDEA is: $\lim_{n \to \infty }a_n=3$, so by the definition of the limit we can say: $\forall \epsilon>0 \exists \bar{n}: |\frac{3n+1}{n}-3|<\epsilon$.
Now we recall the following property of the infimum of a generic $A\subseteq \mathbb R$:
Let $A\subseteq\mathbb R$ is bounded from below, $l=infA\iff\left.\Bigg\lbrace\begin{aligned} & (i) l\leq a\enspace \forall a\in A \\ & (ii) \forall \epsilon>0\enspace \exists\enspace a \in A: l+\epsilon>a \\ \end{aligned} \right.$
The definition of limit tell us that: $-\epsilon<\frac{3n+1}{n}-3<\epsilon\Longrightarrow 3+\epsilon>\frac{3n+1}{n}$. Therefore from the property above we conclude that $infS=3$.
It's all right?

Answer 1

Use the definition of the infimum. Let $\inf S =x$ then by definition for all $\epsilon>0$, one needs to find a $y\in S$ such that $y < x+\epsilon.$ In other words you need to find $n\in \mathbb{N}$ such that $a_n=\frac{3n+1}{n}<x+\epsilon.$ Our guess for the infimum is $x=3$ and so we need to find an $\epsilon>0$ that works. So we try, $\frac{3n+1}{n}<3+\epsilon\implies \frac{1}{n}<\epsilon$ and so if we choose any $n>\frac{1}{\epsilon}$ then we have that $a_n<3+\epsilon$ and thus $\inf S = 3.$