On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like 
$$
a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},
$$
where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.
It gave me
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}
$$
but for me it looks like a blind valley. I would appreciate every help, thank you.

Let
$$a_n=\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=\sqrt[3]n\left(\sqrt[3]{n^2+1}-\sqrt[3]{n^2-1}\right)$$
Defining $f(x)=\sqrt[3]x$ we can apply Mean Value Theorem in $[n^2-1,n^2+1]$.
$$f'(x)=\frac1{3\sqrt[3]{x^2}}$$
For some $\xi\in[n^2-1,n^2+1]$, we have
$$a_n=\frac23\sqrt[3]{\frac n{\xi^2}}\ge\frac23\sqrt[3]{\frac n{(n^2+1)^2}}\ge\frac23\sqrt[3]{\frac n{4n^4}}\ge\frac1{3n}$$
