Convergence of Infinite Series

You can do that as Ron Gordon says, which was also my first hint. So here is a slightly less low tech approach using Taylor expansions. $$ \sqrt[3]{n^3+1}-n=n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)=n\left( 1+\frac{1}{3}\frac{1}{n^3}+O\left(\frac{1}{n^6}\right) -1\right) $$ $$=\frac{1}{3}\frac{1}{n^2}+O\left(\frac{1}{n^5}\right)\sim \frac{1}{3n^2}. $$

So the series converges by limit comparison with $\sum_{n\geq 1}\frac{1}{n^2}$.

Note: I have simply used the Taylor expansion $(1+u)^\alpha=1+\alpha u+O(u^2)$ as $u$ tends to $0$.

Use the relation

$$a^3-b^3=(a-b)(a^2+a b+b^2)$$

with $a=(x^3+1)^{1/3}$ and $b=x$.

The summand is equal to

$$\frac{1}{(x^3+1)^{2/3} + x (x^3+1)^{1/3} + x^{2}} \sim\frac{1}{3 x^2}$$

as $x \rightarrow \infty$. Then use the comparison test.