\(n\in N\)：試證明：\(\displaystyle \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\leq 2\sqrt{n}\)

利用\(\displaystyle \frac{1}{\sqrt{k}}<\frac{2}{\sqrt{k}+\sqrt{k-1}} \)

感恩  我試看看