Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+..+\dfrac{1}{(2n+1)^2}$
$\rightarrow A<\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..+\dfrac{1}{(2n-1).(2n+1)}$
$\rightarrow A<\dfrac{1}{2}.(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{(2n-1).(2n+1)})$
$\rightarrow A<\dfrac{1}{2}.(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+..+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$
$\rightarrow A<\dfrac{1}{2}.(1-\dfrac{1}{2n+1})$
mà $2n+1\ge 2.n\ge 2\rightarrow 1-\dfrac{1}{2n+1} \ge 1-\dfrac{1}{2}=\dfrac{1}{2}$
$\rightarrow A<\dfrac{1}{4}$
