Ta có:
\(\frac{1}{\sqrt{k}}=\frac{2}{2\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}< \frac{2}{\sqrt{k}+\sqrt{k-1}}=\frac{2\left(\sqrt{k}-\sqrt{k-1}\right)}{\left(\sqrt{k}-\sqrt{k-1}\right)\sqrt{k}+\sqrt{k-1}}\)
\(=\frac{2\left(\sqrt{k}-\sqrt{k-1}\right)}{k-\left(k-1\right)}=2\left(\sqrt{k}-\sqrt{k-1}\right)\)