Chứng minh
[TEX]1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{2001}}>86[/TEX]
Mọi người giúp mình nha.

[TEX]A= 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{2001}}[/TEX]
[TEX]=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{ 2}{2\sqrt{2001}}[/TEX]
[TEX]=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{ \sqrt {2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+ \frac {2}{\sqrt{2001}+\sqrt{2001}}[/TEX]
[TEX]A > B= 2(\frac{1}{\sqrt{1}+ \sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+...+\frac{1}{\sqrt{2001}+\sqrt{2002}})[/TEX]
[TEX]B=2(\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2002}-\sqrt{2001}}{2002-2001})[/TEX]
[TEX]=2(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2002}-\sqrt{2001})[/TEX]
[TEX]=2(\sqrt{2002}-1) \approx 87,48 >86[/TEX]
\Rightarrow A>86