Đặt `H = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2`

Ta có: `1/2^2 = 1/2.2 = 1/4`

           `1/3^2 = 1/3.3 < 1/2.3`

           `1/4^2 = 1/4.4 < 1/3.4`

           ...

           `1/100^2 = 1/100.100 < 1/99.100`

`=> H < 1/4 + 1/2.3 + 1/3.4 + ... + 1/99.100`

`=> H < 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100`

`=> H < 1/4 + 1/2 - 1/100`

`=> H < 3/4 - 1/100 < 3/4`

`=> H < 3/4`   (đpcm)

Chúc bạn học tốt!

$\text{ Đặt A=}$${\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2}+...+\dfrac{1}{100^2}}$
$\text{ Ta thấy}$
${ \dfrac{1}{2^2}=\dfrac{1}{2.2}=\dfrac{1}{4}}$
${ \dfrac{1}{3^2}=\dfrac{1}{3.3} < \dfrac{1}{2.3}}$
${ \dfrac{1}{4^2}=\dfrac{1}{4.4} < \dfrac{1}{3.4}}$
....
${\dfrac{1}{100^2}=\dfrac{1}{100.100} < \dfrac{1}{99.100}}$

⇒ ${\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2}+...+\dfrac{1}{100^2}}$ < ${\dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}$

⇒ ${A}$ < ${\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}}$

⇒ ${A}$ ${< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}}$

⇒ ${A}$ ${< \dfrac{3}{4} - \dfrac{1}{100}}$

⇒ ${A}$${ < \dfrac{3}{4}}$