Given the following series: $\sum_{n=1}^{\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha}$ where $\alpha \in \mathbb{R}$. Does the series converge or diverge?
Attempts to solve the problem:
1) $\lim_{n\to\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha } = 0$ - not helpful.
2) Used the formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ - not helpful.
3) The ration test is not helpful either.
The formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ actually can be quite helpful. Note that it implies that
$$\sqrt[3]{n+1}-\sqrt[3]{n-1}=\frac{2}{(n+1)^{2/3}+(n^2-1)^{1/3}+(n-1)^{2/3}}.$$
For sufficiently large $n$ ($n\gt2$ will do for sure, and you can get tighter bounds by going further), that denominator is squeezed between $\frac{3}{2}n^{2/3}$ and $6n^{2/3}$, and therefore the convergence question is the same as for the series $\sum_{n=1}^\infty n^{-2\alpha/3}$.