Ta có:
$\dfrac{1}{2^2}=\dfrac{1}{2.2}<\dfrac{1}{1.2} =\dfrac{2-1}{1.2}=1 -\dfrac12$
$\dfrac{1}{3^2}=\dfrac{1}{3.3}<\dfrac{1}{2.3}=\dfrac{3-2}{2.3}=\dfrac12-\dfrac13$
$\cdots$
$\dfrac{1}{10^2}=\dfrac{1}{10.10}<\dfrac{1}{9.10}=\dfrac{10 -9}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}$
Cộng vế theo vế ta được:
$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{2}+\dfrac12-\dfrac{1}{3}+\cdots+\dfrac19 -\dfrac{1}{10}$
$\Leftrightarrow \dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{10} < 1$
Vậy $\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1$