Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\);\(\dfrac{1}{3^2}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\);.......;\(\dfrac{1}{100^2}< \dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=1-\(\dfrac{1}{100}=\dfrac{99}{100}< 1\)
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)+...+\(\frac{1}{99}-\frac{1}{100}\)
=1-\(\frac{1}{100}\)<1
Vậy \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{100^2}\)<1
