It is given that for $n\in N$

$f_n={\left(n+1\right)}^{1/3}-n^{1/3}$

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

$=\frac{1}{{\left(n+1\right)}^{2/3}+{\left(n+1\right)}^{2/3}{n}^{2/3}+n^{2/3}}$

$\because \forall n\in N$

$3n^{2/3}<{\left(n+1\right)}^{2/3}+{\left(n+1\right)}^{2/3}{n}^{2/3}+n^{2/3}<3{\left(n+1\right)}^{2/3}$

$\Rightarrow \frac{1}{3{\left(n+1\right)}^{2/3}}<\frac{1}{+{\left(n+1\right)}^{2/3}+{\left(n+1\right)}^{2/3}{n}^{2/3}+n^{2/3}}<\frac{1}{3n^{2/3}}$

$\Rightarrow \frac{1}{3{\left(n+1\right)}^{2/3}}<f_n<\frac{1}{3n^{2/3}}$

Similarly,

$\frac{1}{3{\left(n+2\right)}^{2/3}}<f_{n+1}<\frac{1}{3{\left(n+1\right)}^{2/3}}$

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

So, set $A=N$