Cho $A_n=\frac{1}{\left(2n+1\right)\sqrt{2n-1}}$ với $n∈N*$. CMR: $A_1+A_2+A_3+\cdots+A_n<1$

Bùi Ngọc Diễm

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

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

Vì $\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$ $>0$ và $\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$ $<\frac{2}{\sqrt{2n-1}}$

Nên $A_n<$ $\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$

Do đó: $A_1+A_2+A_3+\cdots+A_n<1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}$

⇒ $A_1+A_2+A_3+\cdots+A_n<1-\frac{1}{\sqrt{2n+1}}<1$