<s>[INST] 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?
 [/INST] 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 $|a_n|=\frac{1}{\sqrt{n}+\sqrt{n-1}}$.
</s>