Cho $A=\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+...+\frac{1}{2021^{2}}$. Chứng tỏ rằng $A < \frac{1}{4}$.

Đáp án:

$A=\dfrac1{3^2}+\dfrac1{5^2}+\dfrac1{7^2}\ +\,.\!.\!.+\ \dfrac1{2021^2}\\\Rightarrow A=\underbrace{\dfrac1{2^2}+\dfrac1{3^2}\ +\,.\!.\!.+\ \dfrac1{2021^2}}_{\large A'}-\underbrace{\left(\dfrac1{2^2}+\dfrac1{4^2}\ +\,.\!.\!.+\ \dfrac1{2020^2}\right)}_{\large A''}\\\Rightarrow A'=\dfrac1{2^2}+\dfrac1{3^2}\ +\,.\!.\!.+\ \dfrac1{2021^2}\\\Rightarrow A'<\dfrac1{1.2}+\dfrac1{2.3}\ +\,.\!.\!.+\ \dfrac1{2020.2021}\\\Rightarrow A'<1-\dfrac12+\dfrac12-\dfrac13\ +\,.\!.\!.+\ \dfrac1{2020}-\dfrac1{2021}\\\Rightarrow A'<1-\dfrac1{2021}\\A''=\dfrac1{2^2}+\dfrac1{4^2}\ +\,.\!.\!.+\ \dfrac1{2020^2}\\\Rightarrow A''<\dfrac14+\dfrac1{2.4}\ +\,.\!.\!.+\ \dfrac1{2018.2020}\\\Rightarrow A''<\dfrac14+\dfrac12-\dfrac14\ +\,.\!.\!.+\ \dfrac1{2018}-\dfrac1{2020}\\\Rightarrow A''<\dfrac14+\dfrac12-\dfrac1{2020}\\\Rightarrow A''<\dfrac34-\dfrac1{2020}\\\Rightarrow A=A'-A''<1-\dfrac1{2021}-\left(\dfrac34-\dfrac1{2020}\right)\\\Rightarrow A<1-\dfrac34-\dfrac1{2021}+\dfrac1{2020}\\\Rightarrow A<\dfrac14-\left(\dfrac1{2021}-\dfrac1{2020}\right)\approx \dfrac14\\\Rightarrow A<\dfrac14$

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