TITLE: Test the convergence of a series
QUESTION [7 upvotes]: To test the convergence of a series:
$$
\sum\left[\sqrt[3]{n^3+1}-n\right]
$$
My attempt:
Take out $n$ in common: $\displaystyle\sum\left[n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\right]$.
So this should be divergent.
But, the given answer says its convergent.

REPLY [2 votes]: This is quite similar to the other answers, but slightly different derivation of the estimate:
$$n^3+1 \le n^3+3 +\frac3{n^3}+\frac1{n^6}=\left(n+\frac1{n^2}\right)^3$$
$$\sqrt[3]{n^3+1} \le n+\frac1{n^2}$$
$$\sqrt[3]{n^3+1} -n \le \frac1{n^2}$$
Now we can use comparison test.
If you prefer to have  a tighter estimate, you can use $\left(n+\frac1{3n^2}\right)^3$ instead.