# Series $\sum_{n=1}^{\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha}$ converge or diverge?

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 $\sum_{n=1}^\infty n^{-2\alpha/3}$.
• As $\alpha \in \mathbb{R}$, there isn't a conclusive answer, am I wrong? – Ma.H Nov 27 '10 at 11:02
• But if you know when $\sum_{n=1}^\infty \frac{1}{n^p}$ converges, this shows how you can easily find which values of $\alpha$ make the series converge. The "conclusive answer" will be of the form "the series converge when $\alpha$ is such that ... and diverges otherwise." I'll let you fill in the ellipsis. – Jonas Meyer Nov 27 '10 at 11:05