# Find the infimum and supremum (if any) of set $S=\left\{\frac{n-1}{n+1} | n\geq 1\right\}$

$$S=\left\{\frac{n-1}{n+1} | n\geq 1\right\}$$

I can see that $$0$$ is a lower bound but I want to prove that is an infimum. So I claim that there is $$\epsilon>0$$ where the $$\inf(S)=\epsilon$$.

How do I go about it from here?

$$0$$ is a lower bound and this value is attained when $$n=1$$. So $$0$$ is the infimum and also the minimum of the set.
$$1$$ is an upper bound and no number $$x$$ less than $$1$$ can be an upper bound : $$\frac {n-1} {n+1 } >x$$ when $$n >\frac {1+x} {1-x}$$. Hence $$1$$ is the suprmum of the set but the set does not have a maximum.
• is there a way to prove that $0$ is the infimum algebraically like what was done with the supremum? or is saying that $0\in S$ when $n=1$ enough? – PLC Mar 24 at 9:06
• Saying that $0=\frac {n-1} {n+1} \in S$ when $n=1$ is enough. @PLC – Kavi Rama Murthy Mar 24 at 9:09
Writting $$\frac{n+1-1-1}{n+1} = \frac{n+1 - 2}{n+1} = 1 - \frac{2}{n+1}$$ You see that the sequence is increasing so the inf is the firt term $$0$$ and the sup is the last term $$1$$.
We have $$0 \le s$$ for all $$s \in S$$ and $$0 \in S.$$