In order to prove, that the series $a_n = \sqrt[3]{n+1} - \sqrt{n}$ is monotone, I tried to prove $a_n > a_{n+1} ~\forall n \in \mathbb{N}_{\geq 1}$. I want to know, if there is a neat, less chaotic approach than mine. My approach goes as follows: $$ a_n > a_{n+1} \iff \sqrt[3]{n+1} - \sqrt{n} > \sqrt[3]{n+2} - \sqrt{n+1} \iff \sqrt[3]{n+1} - \sqrt[3]{n+2} > \sqrt{n} - \sqrt{n+1} $$ Now using that $a^2 - b^2 = (a+b)(a-b)$ and $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, we obtain
$$ \iff \frac{(n+1) - (n+2)}{(n+1)^{\frac{2}{3}} + ((n+1)(n+2))^{\frac{1}{3}} + (n+2)^{\frac{2}{3}}} > \frac{n - (n+1)}{\sqrt{n} + \sqrt{n+1}} $$ $$ \iff \frac{-1}{(n+1)^{\frac{2}{3}} + ((n+1)(n+2))^{\frac{1}{3}} + (n+2)^{\frac{2}{3}}} > \frac{-1}{\sqrt{n} + \sqrt{n+1}} $$ $$ \iff \sqrt{n} + \sqrt{n+1} < (n+1)^{\frac{2}{3}} + ((n+1)(n+2))^{\frac{1}{3}} + (n+2)^{\frac{2}{3}} $$ The last equality hold true for all positive n because all root functions are monotone $(*)$ and so we have for all positive n that $$ (n+2)^{\frac{2}{3}} \overset{(*)}{\geq} (n+1)^{\frac{2}{3}} \geq \sqrt{n+1} \overset{(*)}{\geq} \sqrt{n} $$
If some calculus is allowed, $\,\sqrt[3]{n+2} - \sqrt[3]{n+1}=\dfrac{1}{3\sqrt[3]{a^2}}\,$ for some $\,a \in (n+1,n+2)\,$ by the mean value theorem, and $\,\sqrt{n+1} - \sqrt{n} = \dfrac{1}{2 \sqrt{b}}\,$ for some $\,b \in (n,n+1)\,$. It then follows that:
$$ \sqrt[3]{n+2} - \sqrt[3]{n+1} = \dfrac{1}{3\sqrt[3]{a^2}} \lt \underbrace{\dfrac{1}{3\sqrt[3]{(n+1)^2}} < \dfrac{1}{2 \sqrt{n+1}}}_{\text{because}\; n+1 \,\gt\, 1 \,\gt\, (2/3)^6 \;\text{}} \lt \dfrac{1}{2 \sqrt{b}} = \sqrt{n+1} - \sqrt{n} $$