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 $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$
 A: Let $y=1/n$, then $\lim_{n\to \infty}{(n^3+1)}^{1/3}-n=\lim_{y\to0^+}\frac{(1+y^3)^{1/3}-1}{y}$. Using L'Hopital's Rule, this limit evaluates to $0$.Hence, the expression converges.
You can also see that as $n$ increases, the importance of $1$ in the expression ${(n^3+1)}^{1/3}-n$ decreases and $(n^3+1)^{1/3}$ approaches $n$.Hence, the expression converges to $0$(as verified by limit).
  For the series part as @siminore pointed out,this difference ${(n^3+1)}^{1/3}-n$ is of the order of $1/n^2$,therefore, the sum of these is of the order of this sum :$\sum_{1}^{\infty}1/n^2$ which  is $= {\pi}^2/6$. Thus the series is bounded and hence converges .
A: 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.
A: 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)
