Đk: \(n\ge0\)
\(\sqrt{n+2}-\sqrt{n+1}=\frac{1}{\sqrt{n+2}+\sqrt{n+1}}\)
\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
vì \(\sqrt{n+2}+\sqrt{n+1}>\sqrt{n+1}+\sqrt{n}\) <=> \(\frac{1}{\sqrt{n+2}+\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}\)
<=> \(\sqrt{n+2}-\sqrt{n+1}< \sqrt{n+1}-\sqrt{n}\)