Chủ đề này có 1 trả lời

[doctor lee]

choA_n  =$\frac{1}{(2n+1)\sqrt{2n-1}}$

vs n thuoc N *

CM A₁ +A₂ +A₃+...+A_n <1

[NhocThienbinh]

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

Áp dụng vào bài ta sẽ cm đc