Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1}\right)-\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\sum_{k=1}^{n}\left(\sqrt{k+2}-\sqrt{k+1}\right)-\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\left(\sqrt{n+2}-\sqrt{2}\right)-\left(\sqrt{n+1}-1\right)\right]\\
&=\lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n+2}}+1-\sqrt{2}\right)\\
&=1-\sqrt{2}.
\end{align*}
But how to determine the convergence directly by comparison and without evaluating?

MVT:
$A:=\sqrt{n+2}-\sqrt{n+1}=$
$ (1/2)\dfrac{1}{√s},$ $n+1 <s <n+2$;
$B:= \sqrt{n+1}-√n= (1/2)\dfrac{1}{√t}$, $n <t <n+1$;
MVT:
$A-B=(1/2)(-1/2)\dfrac{1}{r^{3/2}}(s-t)$, $t <r <s$;
Finally :
$|A-B| < (1/4)\dfrac{1}{n^{3/2}} \cdot 2$, since
$r >n$, and $s-t <2$.
Comparison test.
