Question

Let $N$ be set of positive integers. For all $n\in N,$ let $f_n=(n+1{)}^{1/3}-n^{1/3}$ and $A=\{n\in N;f_{n+1}<\frac{1}{3(n+1{)}^{2/3}}<f_n\}$ then

• A. $A=N$
• B. $A$ is a finite set
• C. the complement of $A$ in $N$ is nonempty, but finite
• D. $A$ and its complement in $N$ are both finite

Answer 1

The correct option is A $A=N$

$f_n=(n+1{)}^{1/3}-n^{1/3}\Rightarrow f_n=\frac{1}{(n+1{)}^{2/3}+n^{2/3}+n^{1/3}(n+1{)}^{1/3}}$

Now, For all $n\in N$
$n^{2/3}<(n+1{)}^{2/3}$
$n^{1/3}(n+1{)}^{1/3}<(n+1{)}^{2/3}$
$\Rightarrow (n+1{)}^{2/3}+n^{2/3}+n^{1/3}(n+1{)}^{1/3}<3(n+1{)}^{2/3}\Rightarrow \frac{1}{(n+1{)}^{2/3}+n^{2/3}+n^{1/3}(n+1{)}^{1/3}}>\frac{1}{3(n+1{)}^{2/3}}\Rightarrow f_n>\frac{1}{3(n+1{)}^{2/3}}....(1)$

Moreover, For all $n\in N$
$(n+2{)}^{2/3}>(n+1{)}^{2/3}$
$(n+2{)}^{1/3}(n+1{)}^{1/3}>(n+1{)}^{2/3}$
$\Rightarrow (n+2{)}^{2/3}+(n+1{)}^{2/3}+(n+1{)}^{1/3}(n+2{)}^{1/3}>3(n+1{)}^{2/3}\Rightarrow \frac{1}{(n+2{)}^{2/3}+(n+1{)}^{2/3}+(n+1{)}^{1/3}(n+2{)}^{1/3}}<\frac{1}{3(n+1{)}^{2/3}}\Rightarrow f_{n+1}<\frac{1}{3(n+1{)}^{2/3}}....(2)$

From equation $\left(1\right)$ and $\left(2\right)$,
$f_{n+1}<\frac{1}{3(n+1{)}^{2/3}}<f_n,\forall n\in N$