Let $(z_n)$ be the sequence given by $$z_n = (−1)^n + \frac{1}{n} \qquad n \in \{1, 2, 3, \cdots\}$$
Let $A\subset R$ be a subset with the properties: $\inf A = −2$ and $\sup A = 1$.
Let $$B := \{ a \cdot z_n \mid a \in A, n \in\Bbb{N}_+\}$$
Find $\sup B$.
We know that$$\large z_{2n}=1+\frac{1}{2n}\\z_{2n-1}=-1+\frac{1}{2n-1}$$therefore$$\large\sup_{n}\{z_n\}=\max\{\sup_{2n}\{z_{2n}\},\sup_{2n-1}\{z_{2n-1}\}\}=\max\{z_2,-\frac{1}{2}\}=z_2=\frac{3}{2}\\\large\inf_{n}\{z_n\}=\min\{\inf_{2n}\{z_{2n}\},\inf_{2n-1}\{z_{2n-1}\}\}=\min\{1,-1\}=-1$$therefore we have that:$$\forall a\in A,n\to -1<z_n\le \frac{3}{2}\quad ,\quad -2\le a\le1\to {a.z_n<2\\a.z_n\le \frac{3}{2}}$$which leads us to $$\forall b\in B\quad,\quad b<2$$which implies that $l=2$ is an upper bound for $B$. To prove that is least such bound it suffices to prove that for any $\epsilon>0$ there exists a $b\in B$ such that $b>2-\epsilon$. Now take $b=az_n$ for some $n$ and $a\in A$. Since $\inf A=-2$ we can get arbitrarily close to it i.e. $a=-2+\epsilon_1$ when $\epsilon_1>0$ and arbitrary. Therefore:$$b=a.z_n=(-2+\epsilon_1)(-1+\frac{1}{2n-1})=2-\epsilon_1-\frac{2}{2n-1}+\frac{\epsilon_1}{2n-1}$$ By choosing $n>\frac{2}{\epsilon}+\frac{1}{2}$ and $\epsilon_1=\frac{\epsilon}{2}$ we get:$$2-\epsilon_1-\frac{2}{2n-1}+\frac{\epsilon_1}{2n-1}>2-\epsilon_1-\frac{2}{2n-1}>2-\frac{\epsilon}{2}-\frac{\epsilon}{2}=2-{\epsilon}$$which is what we wanted to show. Then:$$\sup B=2$$