Testing the convergence of cube root of some function of n I have to test the convergence of the following series:-
$$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$
My approach is as follows :-
$$n^3+1>1=\sqrt[3]{n^3+1}>1=\sqrt[3]{n^3+1}-n>1-n$$
Now since$\sum 1-n$ diverges, the series under consideration diverges.
Is this right or wrong?

Hint
Let $a=\sqrt[3]{n^3+1}$ and $b=n$. Then,
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{1}{a^2+ab+b^2}.$$
Observe that $a \geq b$, therefore 
$$\sqrt[3]{n^3+1}-n=a-b \leq \frac{1}{3b^2}=\frac{1}{3n^2}.$$
Now use comparison to claim convergence.
