Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$ I want to expand and test this $\{(n^3+1)^{1/3} - n\}$ for convergence/divergence.
The edited version is: Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$

By direct inspection, for every pair of real numbers $A$ and $B$,
$$
A^3 - B^3 = (A-B) (A^2+AB+B^2).
$$
Choose now $A=\sqrt[3]{n^3+1}$ and $B=n$. Then
$$
(n^3+1)^{1/3} - n = \frac{n^3+1-n^3}{(n^3+1)^{2/3} + n (n^3+1)^{1/3}+n^2} \sim \frac{1}{n^2}
$$
as $n \to +\infty$. The limit is therefore zero.
