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 $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$

The first thing we note is that, as $n\to \infty$ the term $(n^3+1)^{1/3}$ is of order $n$, which is cancelled by the other term, hence we must ask about the next order approximation.
Now, $ (n^3+1)^{1/3} = n \left(1 + \frac{1}{n^3}\right)^{1/3}$ and $(1+x)^{1/3} = 1 + \frac{1}{3}x + O(x^2)$, hence
$$ (n^3+1)^{1/3} - n = \frac{1}{3}\frac{1}{n^2}+O(n^{-5})$$
Hence, the series converge.
(If you want an explicit bound, you can just use $ (1+x)^{1/3} < 1+x$ for $x>0$, so you can compare to the series $\sum 1/n^2$ which you're supposed to know is convergent)
