Find the sum $\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ $$\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$$
On their own, all three are divergent, so I thought the best way would be to rewrite it as:
$$\frac{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}\cdot \left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)=\frac{\left(\left(\sqrt{n+2}-2\sqrt{n+1}\right)^2-n\right)}{\left(\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}\right)}$$
But that doesn't really make anything simpler.

Render the summand as $(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})$ and use telescoping.  The difference between two consecutive square roots goes to zero as the arguments go to infinity.  You should be able to see the answer immediately.
