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.

Hint We can apply a cubic analogue of the technique usually called multiplying by the conjugate, which itself is probably familiar from working with limits involving square roots. In this case, write $$u = \sqrt[3]{n^3 + 1},$$ so that $$u^3 - n^3 = (n^3 + 1) - (n^3) = 1.$$
On the other hand, factoring gives (in analogy to the difference of perfect squares factorization $a^2 - b^2 = (a - b)(a + b)$ used in the square root case case) $$u^3 - n^3 = (u - n) (u^2 + un + n^2).$$

 Explicitly, multiplying gives that we can write our sum as \begin{align} \sum_{n = 0}^{\infty} (u - n) &= \sum_{n = 0}^{\infty} (u - n) \cdot \frac{u^2 + un + n^2}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{u^3 - n^3}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{1}{u^2 + un + n^2} . \end{align} Now, $u > n$, and so the summand satisfies $$\frac{1}{u^2 + un + n^2} < \frac{1}{3 n^2}.$$ The given series is bounded by $$\frac{1}{3} \sum_{n = 0}^{\infty} \frac{1}{n^2},$$ which converges by, e.g., the $p$-test.

