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Let $N$ be the set of positive integers. For all $n \in N$, let $f_n=(n+1)^{1 / 3}-n^{1 / 3} \text { and }$ $A=\left\{n \in N: f_{n+1}<\frac{1}{3(n+1)^{2 / 3}} < f_n\right\}$ Then,
$A=N$
$A$ is a finite set
the complement of $A$ in $N$ is nonempty, but finite
$A$ and its complement in $N$ are both infinite
Solution
(a)
It is given that for $n \in N$
$f_n=(n+1)^{1 / 3}-n^{1 / 3}\,(n+1)-n$
$(n+1)^{2/3}+(n+1)^{23} n^{2/3}+n^{2/3}$
$3 n^{2 / 3} < (n+1)^{2 / 3}+(n+1)^{2/3} n^{2/3}+n^{2/3} < 3(n+1)^{2 / 3}$
$\Rightarrow \frac{1}{3(n+1)^{2/3}} < \frac{1}{(n+1)^{2 / 3}+(n+1)^{2 / 3} n^{2/3}+n^{2/3}} < \frac{1}{3 n^{2/3}}$
$\frac{1}{3(n+1)^{2/3}} < f_n < \frac{1}{3 n^{2/3}}$
Similarly,
$\frac{1}{3(n+2)^{23}} < f_{n+1} < \frac{1}{3(n+1)^{23}}$
$\therefore \quad f_{n+1} < \frac{1}{3(n+1)^{23}} < f_{n+1}, \forall n \in N$
So, $\operatorname{set} A=N$.
