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?
 A: 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.
A: It’s actually convergent. Treat your general term $\sqrt[3]{n^3+1}-n$ as a fraction with $1$ underneath, and then multiply top and bottom by
$\bigl(\sqrt[3]{n^3+1}\,\bigr)^2+n\sqrt[3]{n^3+1}+n^2$. Use the fact that $\sqrt[3]{n^3+1}>n$ as well.
A: 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.
A: Similar to André Nicolas's more elementary solution, but this method is perhaps more widely applicable. Use the Taylor expansion $(1 + t)^{1/3} = 1 + t/3 + o(t)$ near $t = 0$.
Call the general term of the series $a_n$. We have $a_n = \sqrt[3]{n^3 + 1} - n = n\left[(1 + 1/n^3)^{1/3} - 1 \right] = n[1/3n^3 + o(1/n^3)] = 1/3n^2 + o(1/n^2).$ Thus $a_n \sim 1/3n^2$ as $n \to +\infty$. Therefore $\sum a_n$ converges by comparison with $\sum 1/n^2$.
