Investigate convergence of a series using comparison test

Question

The series with $a_n$:

$a_n = (n^{1/3}-(n-1)^{1/3})/n^{1/2} $

I tried comparing it to the $1/n^2$ and $1/n^{3/2}$ because those definitely converge, but proving the inequality gives rise to pretty complicated polynomials with high degree, and I cant seem to find an elegant proof that this series converges.

Answer 1

HINT:

Using $\displaystyle a-b=\frac{a^3-b^3}{a^2+ab+b^2}$ with $a=n^{1/3}$ and $b=(n-1)^{1/3}$, we have for $n\ge 1$

$$\begin{align} n^{1/3}-(n-1)^{1/3}&=\frac{1}{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}\\\\ &\le \frac{1}{n^{2/3}} \end{align}$$