[tex]\displaystyle\bf\\\text{Folosim formula: }~~~~~\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\\\\text{Rezolvare:}\\\\\\S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{97\cdot98}+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\\\\\\S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\\\\\text{\bf Se reduc termenii asemenea.}\\\\S=\frac{1}{1}-\frac{1}{100}=\boxed{\bf\frac{99}{100}<1}[/tex]
