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?

Note that
$$\sqrt[3]{n^3+1}-n=n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right).$$
For positive $t$ we have 
$$1\lt \sqrt[3]{1+t}\lt 1+\frac{1}{3}t,$$
since $\left(1+\frac{t}{3}\right)^3\gt 1+t$. It follows that
$$0\lt n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\lt \frac{1}{3n^2}$$
and now convergence of our series follows by comparison.
