转：一个有理化不等式放缩

realnumber

湖南颜文杰

通分，有理化bn<1/(2(n+2)(n+1))

realnumber

\[b_n=\frac{\sqrt{n}}{(n+1)\sqrt{n+2}}((n+1)-\sqrt{n(n+2)})=\frac{\sqrt{n}}{(n+1)\sqrt{n+2}}\frac{1}{(n+1)+\sqrt{n(n+2)}}<\frac{\sqrt{n}}{(n+1)\sqrt{n+2}}\frac{1}{2\sqrt{n(n+2)}}=\frac{1}{2}(\frac{1}{n+1}-\frac{1}{n+2})\]