回復 3# tian 的帖子
\(\begin{align}
& n+1>\sqrt{n+1}\times \sqrt{n} \\
& 2\left( n+1 \right)>\left( n+1 \right)+\sqrt{n+1}\times \sqrt{n} \\
& n+1>\frac{\left( n+1 \right)+\sqrt{n+1}\times \sqrt{n}}{2} \\
& \left( n+1 \right)\sqrt{n}>\frac{\sqrt{n}\left[ \left( n+1 \right)+\sqrt{n+1}\times \sqrt{n} \right]}{2}=\frac{\sqrt{n+1}\times \sqrt{n}\left( \sqrt{n+1}+\sqrt{n} \right)}{2}=\frac{\sqrt{n+1}\times \sqrt{n}}{2\left( \sqrt{n+1}-\sqrt{n} \right)} \\
& \frac{1}{\left( n+1 \right)\sqrt{n}}<\text{}\frac{2\left( \sqrt{n+1}-\sqrt{n} \right)}{\sqrt{n+1}\times \sqrt{n}}\text{=2}\left( \frac{\text{1}}{\sqrt{n}}-\frac{\text{1}}{\sqrt{n+1}} \right) \\
\end{align}\)
