ta có \frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18
ta có $\frac{1}{\sqrt{k}} = \frac{2}{\sqrt{k} + \sqrt{k}} \geq \frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k}\Rightarrow\frac{1}{\sqrt{2}}...>17tương tự \frac{1}{k}\leq2(\sqrt{k}-\sqrt{k-1} \Rightarrow\frac{1}{\sqrt{2}}...<18$