Cho $A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}$
Chứng minh $\frac{87}{89}<A<\frac{88}{45}$
Cho $A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}$
Chứng minh $\frac{87}{89}<A<\frac{88}{45}$
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 $A<2\left [ \left ( \frac{1}{\sqrt{1}} -\frac{1}{\sqrt{2}}\right ) +\left ( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}} \right )+...+\left ( \frac{1}{\sqrt{2010}} -\frac{1}{\sqrt{2011}}\right )\right ]$
$=2\left ( 1-\frac{1}{\sqrt{2011}} \right )<2\left ( 1-\frac{1}{\sqrt{2025}} \right )=\frac{88}{45}$
"God made the integers, all else is the work of man."
Leopold Kronecker
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