Convergence test for a series

Question

Let S = $\sum_{n=1}^{\infty}{\frac{n^k}{((n^3+n)^{\frac{1}{3}}-n)}},\forall\space\space\space\space k\in\mathbb{Z}.$ Say if this series converge of diverge.

My attempt:

$$S=\sum_{n=1}^{\infty}\frac{(n^3+n)^{\frac{2}{3}}+n(n^3+n)^{\frac{1}{3}}+n^2}{n^{1-k}}$$

Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $\to0$ iff $1-k<2\implies k>1$ and divergent for $k<1$.

Answer 1

Note that\begin{align}\frac1{\sqrt[3]{n^3+n}-n}&=\frac1{\sqrt[3]{n^3+n}-\sqrt[3]{n^3}}\\&=\frac{\sqrt[3]{(n^3+n)^2}+\sqrt[3]{(n^3+n)n^3}+\sqrt[3]{(n^3)^2}}{n^3+n-n^3}\\&=\frac{\sqrt[3]{(n^3+n)^2}+\sqrt[3]{(n^3+n)n^3}+\sqrt[3]{(n^3)^2}}n\\&\in\left[3n,\frac{3\sqrt[3]{(n^3+n)^2}}n\right].\end{align}This is enough to prove that the series converges if and only if $k<-2$.

Answer 2

May be, you could start considering that, for large values of $n$, $$(n^3+n)^{\frac{2}{3}}+n(n^3+n)^{\frac{1}{3}}+n^2=3 n^2+1-\frac{2}{9 n^2}+O\left(\frac{1}{n^4}\right)$$