We consider the sets \begin{equation*}S_1:=\left \{\frac{2^n}{n}\mid n\in \mathbb{N}\right \} \ \text{ and } \ S_2:=\left \{\frac{1}{n}-2^{-n}\mid n\in \mathbb{N}\right \}\end{equation*} Determine the supremum and the infimum (if possible) and check if these are maximum and minimum respectively.
For the first set, it is not bounded from above, is it?
If we consider the second set do we consider it as a sum of sets $\left \{\frac{1}{n}\right \}$ and $\left \{-2^{-n}\right \}$ ?
Best Answer
The sequence $\frac{2^n}{n}$ is increasing an unbounded, so the infimum (and minimum) is $\frac{2^1}{1}=2$ and there is no supremum/maximum. Regarding the second questions you dot not sum sets...
$S_2$ is a set of numbers obtained by the formula $\frac 1n -2^{-n}$, it has no common elements with the sets you mention. The sequence $\frac 1n -2^{-n}$ is decreasing and tends to zero, so de supremum/maximum is $\frac 11 -2^{-1}= \frac 12$, the infimum is zero, but it is not the minimum.