Convergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$ I have encountered the following problem:

Determine whether $$\sum \limits_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$$ converges or diverges.

What I have tried so far:
Assume that $a_n = \sqrt[3]{n^3+1} - n$.


*

*$n$th term test: $\lim \limits_{x\to\infty} a_n = 0$ $\Rightarrow$ inconclusive.

*$p$-series test: can't be used

*geometric series test: can't be used

*alternating series test: not alternating, can't use

*telescoping series: does not look like it (after rewriting first few terms)

*integral test: integrating the $a_n$ is outside the scope of my course (checked result with wolfram)

*ratio test: simplification of the limit is not possible due to number of expanding terms

*comparison / limit comparison test: no idea what should I compare it to


I would be grateful for any help.

For any $x\in[0,1]$, the inequality:
$$ (1+x)^{\frac{1}{3}}\leq 1+\frac{x}{3} \tag{1}$$
holds by concavity or just by considering the cube of both terms. It follows that:
$$ 0\leq \sqrt[3]{1+n^3}-n = n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\leq\frac{1}{3n^2}\tag{2}$$
hence the given series is convergent by the p-test or by the inequality:
$$ 0\leq \sum_{n=1}^{N}\frac{1}{n^2} \leq 1+\sum_{n=1}^{N-1}\frac{1}{n(n+1)}=2-\frac{1}{N}\leq 2.\tag{3}$$
