What criterion of convergence should I use here?

Question

I'm trying to check if the series $$\sum_{n=1}^{\infty} \sqrt[3]{n^3+n}-n$$ does converge and I don't know which criterion should I use. D'Alembert doesn't help, what's more I can't limit this series with a sense to use comparative criterion. The necessary condition is fullfiled, but it is unsufficient. Thanks for help :)

Answer 1

Hint:

$\sqrt[3]{n^3+n}=n\left(1+\dfrac1{n^2}\right)^{\frac13}$ and try a Taylor expansion.

Answer 2

Proof of divergence without using series expansions: verify that $(n+\frac 1 {4n})^{3} <n^{3}+n$ for all $n$ by expanding $(n+\frac 1 {4n})^{3}$. It follows that $({n^{3}+n})^{\frac 1 3}-n > \frac 1 {4n}$.

Answer 3

From $a^3-b^3=(a-b)(a^2+ab+b^2)$ you get

$$ \sqrt[3]{a} - \sqrt[3]{b} = \frac{{a - b}} {{\sqrt[3]{{a^2 }} + \sqrt[3]{{ab}} + \sqrt[3]{{b^2 }}}} $$ Therefore $$ \begin{gathered} \sqrt[3]{{n^3 + n}} - \sqrt[3]{{n^3 }} = \frac{{n^3 + n - n^3 }} {{\sqrt[3]{{\left( {n^3 + n} \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n} \right)}} + \sqrt[3]{{n^6 }}}} = \hfill \\ \hfill \\ = \frac{n} {{\sqrt[3]{{\left( {n^3 + n} \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n} \right)}} + \sqrt[3]{{n^6 }}}} \geqslant \hfill \\ \hfill \\ \geqslant \frac{n} {{\sqrt[3]{{\left( {n^3 + n^3 } \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n^3 } \right)}} + \sqrt[3]{{n^6 }}}} = \hfill \\ \hfill \\ = \frac{n} {{\sqrt[3]{{4n^6 }} + \sqrt[3]{{2n^3 }} + \sqrt[3]{{n^6 }}}} > \frac{n} {{3n^2 + 3n^2 + 3n^2 }} = \frac{1} {{9n}} \hfill \\ \end{gathered} $$