Find the $\inf$ and the $\sup$ of a set defined by a succession

Question

I have a problem of mathematical analysis that asks me to find the $\inf$ and the $\sup$ of these two following sets defined by a succession: A=$2(-1)^{n+1}-(-1)^\frac{n(n-1)}{2}\times(2+\frac{3}{n}),for n\in N^*$ and B={$\frac{n-1}{n+1}\cos(\frac{2n\pi}{3})$, for n\in N} Where $N^{\ast}$ is the set of natural numbers without the zero. I know that the successions that defines these two sets can be expressed thought four different sub succession for the set $A$ and with 3 sub succession for the set $B$... but how can I arrive formally to these sub successions?

Answer 1

Hint. As regards $A$, the factor $(-1)^{\frac{n(n-1)}{2}}$ (which has period $4$), suggests to consider the subsequences given by $n=4k+r$ for $r=0,1,2,3$.

i) For $r=0$, the term is $\displaystyle -2-\left(2+\frac{3}{4k}\right)=-4-\frac{3}{4k}$ for $k\geq 1$

ii) For $r=1$, the term is $\displaystyle 2-\left(2+\frac{3}{4k+1}\right)=-\frac{3}{4k+1}$ for $k\geq 0$.

iii) For $r=2$, the term is $\displaystyle -2+\left(2+\frac{3}{4k+2}\right)=\frac{3}{4k+2}$ for $k\geq 0$.

iv) For $r=3$, the term is $\displaystyle 2+\left(2+\frac{3}{4k+3}\right)=4+\frac{3}{4k+3}$ for $k\geq 0$.

What may we conclude?

The approach for $B$ is similar, just consider the possible values of the $3$-periodic term $\cos\left(\frac{2n\pi}{3}\right)$.