Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}=\frac{1}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)-\sqrt{n(n+1)}}<\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}-\sqrt{n(n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......
\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cộng theo vế:
\(\Rightarrow \text{VT}< 1-\frac{1}{\sqrt{n+1}}\) (đpcm)