Cần giúp đỡ chứng minh BDT

**dieuuocsaobang** · 07-26-2010, 05:52 PM

CHỨNG MINH BẤT ĐẲNG THỨC (BDT)

Bài 1 $\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{ 1}{(n+1)\sqrt{n}}<2$ mọi n
Bài 2 $\sum_{n}^{k=1}\frac{1}{(2k+1)(2k-1)}<\frac{1}{2}$ mọi n

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**binhdt** · 07-27-2010, 07:41 PM

Nguyên văn bởi dieuuocsaobang

Bài 1 $\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{ 1}{(n+1)\sqrt{n}}<2$ mọi n

$\frac{1}{(n+1)\sqrt{n}} = \frac{\sqrt{n}}{n(n+1)} =\sqrt{n}(\frac{1}{n}-\frac{1}{n+1}) =\sqrt{n}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}+\frac{1}{ \sqrt{n+1}})<2(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$
$=>VT< 2(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-....+ \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})<2$ .