Question

[Hoàng Bảo]

Cho $A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\mathrm{...}+\frac{1}{{100}^2}$. Chứng minh $A<\frac{1}{2}$.

Answer 1

[Nguyễn Lê Phước Thịnh]

\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)

=>\(A=\dfrac{1}{2^2}+\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}+\dfrac{1}{2^2}\cdot\dfrac{49}{50}=\dfrac{1}{4}\left(1+\dfrac{49}{50}\right)=\dfrac{1}{4}\cdot\dfrac{99}{50}=\dfrac{99}{200}< \dfrac{1}{2}\)