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?
 A: 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.
A: \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
$$ \sum_{n=1}^\infty (\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}) \text{ is convergent} \Leftrightarrow \sum_{n=1}^\infty \frac{1}{n^{3/2}} \text{ is convergent}  $$
A: Just an alternative answer and using a different test from
$$\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}=\\
(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})=\\
\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
and using alternating series test the series converges, where $|a_n|=\frac{1}{\sqrt{n}+\sqrt{n-1}}$.
A: 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.
