Đáp án + Giải thích các bước giải:
Ta có:
$\dfrac{1}{2^2} + \dfrac{1}{3^2} +...+ \dfrac{1}{100^2} < \dfrac{1}{1.2} + \dfrac{1}{2.3} +...+ \dfrac{1}{99.100}\\$
$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} +...+ \dfrac{1}{99} - \dfrac{1}{100}\\$
$= 1 - \dfrac{1}{100}\\$
$=\dfrac{99}{100}\\$
$\Rightarrow \dfrac{1}{2^2} + \dfrac{1}{3^2} +... + \dfrac{1}{100^2} < \dfrac{99}{100}\\$
Mà $\dfrac{99}{100} > \dfrac{3}{4}\\$
$\Rightarrow \dfrac{1}{2^2} + \dfrac{1}{3^2} +... + \dfrac{1}{100^2} <\dfrac{3}{4}$
