Ta có
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{100}\)
=>.....