Đáp án:
Giải thích các bước giải:
$\frac{1}{(2n+1)\sqrt{2n-1}}=\frac{\sqrt{2n-1}}{(2n-1)(2n+1)}=\frac{\sqrt{2n-1}}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})$
$A_{n}=\frac{\sqrt{2n-1}}{2}(\frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{2n+1}})(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}) < \frac{\sqrt{2n-1}}{2}(\frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{2n-1}})(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}})$
$A_{n}<\frac{\sqrt{2n-1}}{2}.\frac{2}{\sqrt{2n-1}}.(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}})=\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$
Vậy:
$A_{1}+A_{2}+...+A{n}<\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$
$⇒A_{1}+A_{2}+...+A{n}<1-\frac{1}{\sqrt{2n+1}}<1$ (đpcm)
