Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test

Using $\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}$ repeatedly, we have $$\begin{align}\sqrt{n+2}-2\sqrt{n+1}+\sqrt n&=\left(\sqrt{n+2}-\sqrt{n+1}\right)-\left(\sqrt{n+1}-\sqrt n\right)\\&=\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}} \\&=\frac{\sqrt n-\sqrt{n+2}}{(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+1}+\sqrt{n})} \\&=-\frac{2}{(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+1}+\sqrt{n})(\sqrt n+\sqrt{n+2})}, \end{align}$$ hence $$\left|\sqrt{n+2}-2\sqrt{n+1}+\sqrt n\right|<\frac1{4 n^{3/2}}, $$ which is good enough for the comparison test.

\begin{align} \sqrt{n+2}-2\sqrt{n+1}+\sqrt{n} &= \frac{(\sqrt{n+2}+\sqrt{n})^2-(2\sqrt{n+1})^2}{\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n}} = \\ &= \frac{2(\sqrt{n+2}\sqrt{n}-n-1)}{\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n}} = \\ &= \frac{2\big((\sqrt{n+2}\sqrt{n})^2-(n+1)^2\big)}{(\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n})(\sqrt{n+2}\sqrt{n}+n+1)}= \\ &= \frac{-2}{(\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n})(\sqrt{n+2}\sqrt{n}+n+1)}= \\ &= \frac{1}{n^{3/2}}\frac{-2}{(\sqrt{1+\frac2n}+2\sqrt{1+\frac1n}+1)(\sqrt{1+\frac2n}+1+\frac1n)}\end{align} $$ \lim_{n\rightarrow\infty}\frac{-2}{(\sqrt{1+\frac2n}+2\sqrt{1+\frac1n}+1)(\sqrt{1+\frac2n}+1+\frac1n)} = -\frac14 $$ so, by comparison test $$ \sum_{n=1}^\infty (\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}) \text{ is convergent} \Leftrightarrow \sum_{n=1}^\infty \frac{1}{n^{3/2}} \text{ is convergent} $$