Finding $\sup$,$\inf$,$\max$ $\min$ of set $A=\{x \in \Bbb R | x=\frac{(-1)^{n+1}}{n+1} + \frac{1+(-1)^n}{2}\}$
let $a_n=\frac{(-1)^{n+1}}{n+1} + \frac{1+(-1)^n}{2}$
first I tried putting the sequence as in even and odd indexes and got $a_{2n}=\{\frac{4}{5},\frac{8}{9},\frac{12}{13},\frac{16}{17},...\}$ and $a_{2n-1}=\{\frac{1}{4},\frac{1}{8}\frac{1}{12},\frac{1}{16},...\}$
from here the impression is that $a_n$ is bounded by $0<a_n<1$ and if according to the real numbers completeness axiom if a set is bounded then there exists infimum and supremum
I tried proving $supA=1$ by using archimedean property. assume there exists an $M \in \Bbb R$ such that $a<M<1$ $(a \in A)$ we will pick an $\varepsilon=1-M$ so according to the archimedean property there exists $n_0 \in \Bbb N$ such that $0<\frac{1}{n_0}<\varepsilon$ so we have $\frac{1}{n_0}<1-M$ $\to$ $\frac{1}{n_0}+M<1$ and this is a contradiction so $supA=1$
showing $infA=0$ by the same method assume there exists $n>m>0$ $(n \in A)$ and we pick $\varepsilon=m$ so according to archimedean property $0<\frac{1}{n_0}<m$ and also we get a contradicition so $infA=0$
maximum and minimum do not exists because $0,1$ are not in the set $A$
I feel like my solution are not right , I mostly did how my professor solved a certain question so my solution does not really come from understanding.
I also thought about trying to prove the sequence is monotone and bounded and then the limits would give us $sup$ or $inf$ accordingly but I did not know how to do that here
EDIT - since $a_{2n-1}$ is decreasing , if i calculate the limit will it give me the infimum? and same goes for $a_{2n}$ but then I will get the supremum? because monotone sequence + bounded
Thanks for any help and tips