Answer
$\left\{ {\sqrt[3]{{n + 1}} - n} \right\}$ is decreasing for $n>0$.
Work Step by Step
Let $f$ be a function such that $f\left( x \right) = \sqrt[3]{{x + 1}} - x$. So, $f\left( n \right) = {a_n}$.
The derivative of $f$ is
$f'\left( x \right) = \frac{1}{3}{\left( {x + 1} \right)^{ - 2/3}} - 1 = \frac{1}{{3{{\left( {x + 1} \right)}^{2/3}}}} - 1$
For $x>0$, the term $\frac{1}{{3{{\left( {x + 1} \right)}^{2/3}}}}$ is less than 1. Therefore, $f'\left( x \right) < 0$. So, $f\left( x \right)$ is decreasing.
It follows that $\left\{ {\sqrt[3]{{n + 1}} - n} \right\}$ is decreasing for $n>0$.
