Chủ đề này có 1 trả lời

[yeutoanmaimai1]

Cho $A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}$

Chứng minh $\frac{87}{89}<A<\frac{88}{45}$

[the man]

Cho $A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}$

Chứng minh $\frac{87}{89}<A<\frac{88}{45}$

- Chứng minh $A>\frac{87}{89}$

 Ta có $\frac{1}{(k+1)\sqrt{k}}=\frac{\sqrt{k}}{k(k+1)}=\sqrt{k}.\left ( \frac{1}{k} -\frac{1}{k+1}\right )=\frac{1}{\sqrt{k}}-\frac{\sqrt{k}}{k+1}$

           $>\frac{1}{\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$

 Từ đó $A>\left ( \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}} \right )+\left ( \frac{1}{\sqrt{2}} -\frac{1}{\sqrt{3}}\right )+...+\left ( \frac{1}{\sqrt{2010}} -\frac{1}{\sqrt{2011}}\right )$

        $=1-\frac{1}{\sqrt{2011}}>1-\sqrt{\frac{4}{7921}}=\frac{87}{89}$

-Chứng minh $A<\frac{88}{45}$

 Ta có $\frac{1}{(k+1)\sqrt{k}}=\frac{1}{\sqrt{k(k+1)}}.\frac{2}{2\sqrt{k+1}}$

           $<\frac{1}{\sqrt{k(k+1)}}.\frac{2}{\sqrt{k+1}+\sqrt{k}}$

           $=\frac{2}{\sqrt{k(k+1)}}.(\sqrt{k+1}-\sqrt{k})=2\left ( \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} \right )$

Từ đó $A<2\left [ \left ( \frac{1}{\sqrt{1}} -\frac{1}{\sqrt{2}}\right ) +\left ( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}} \right )+...+\left ( \frac{1}{\sqrt{2010}} -\frac{1}{\sqrt{2011}}\right )\right ]$

              $=2\left ( 1-\frac{1}{\sqrt{2011}} \right )<2\left ( 1-\frac{1}{\sqrt{2025}} \right )=\frac{88}{45}$