Convergence of $\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$ I am trying to test the convergence of
$$\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$$
The ratio test yields and inconclusive $1$. I am able to show $\sqrt{n^2+1} - \sqrt[3]{n^3+1} < 1$ but this is not enough for convergence. One idea would be to rearrange the expression so that it is of the form ${1 \over f(n)}$ and then show that $\text{deg}f > 1$. With the fact that $\sum{1 \over n^s}$ converges if $s>1$ this would give me convergence. So, we could use the formula for $a^3-b^3$ and then the formula for $a^2-b^2$. Is there a quicker way?

$$ \sqrt{n^2-1}-\sqrt[3]{n^3-1} = \left(n-\sqrt[3]{n^3-1}\right)-\left(n-\sqrt{n^2-1}\right)\tag{1} $$
so, due to $\frac{a^2-b^2}{a-b}=(a+b)$ and $\frac{a^3-b^3}{a-b}=(a^2+ab+b^2)$ we have:
$$\left(n-\sqrt[3]{n^3-1}\right)\leq \frac{1}{n^2},\qquad \left(n-\sqrt{n^2-1}\right)\geq \frac{1}{2n}\tag{2} $$
and these inequalities ensure that the given series is divergent.
