I have to prove:
If $x_n=\frac{3n-1}{n},\ x\in\Bbb{R},n \in \Bbb{N}$, then $\displaystyle\inf_{n\in\Bbb{N}}\{x_n\}=3$.
First I have to prove by induction the sequence is growing, but then i got this and I'm stuck:
$\frac{3n+2}{n+1}>\frac{3n-1}{n}$ not sure what to do next.
A proof that $3$ is indeed the supremum, not the infimum:
To prove that the sequence is increasing, consider, it is always true that $$3n^2 + 2n > 3n^2 + 2n -1$$ but we can re-write this inequality as $$(3n + 2)n > (3n-1)(n+1)$$ then cross divide to get $$\frac{3n+2}{n+1} > \frac{3n-1}{n}.$$ Thus the sequence is increasing. The sequence is also bounded above by $3$. Indeed $$\frac{3n-1}{n} \le \frac{3n}{n} = 3.$$ An increasing sequence which is bounded above converges to its supremum. Thus to find the supremum, we simply need to take the limit. We see $$\lim_{n\to \infty} \frac{3n-1}{n} = \lim_{n\to \infty} 3 - \frac{1}{n} = 3.$$ Thus $3$ is the supremum of the sequence. Alternatively, you could argue by definition. We showed above that $3$ is an upper bound. To prove it is the least upper bound, we need to show that $3 - \epsilon$ is not an upper bound for any $\epsilon > 0$. Indeed, if $\epsilon > 0$, then there is $N \in \mathbb N$ with $\frac{1}{N} < \epsilon$. But then $-\frac{1}{N} > - \epsilon$ so $$3-\frac{1}{n} > 3 - \epsilon \,\,\, \Longleftrightarrow \,\,\, \frac{3n-1}{n} > 3-\epsilon.$$ Thus $3-\epsilon$ is not an upper bound for the sequence, so $3$ is the least upper bound, i.e., 3 is the supremum.