We have : 

\(\dfrac{1}{10.11}=\dfrac{1}{10}-\dfrac{1}{11},\dfrac{1}{11.12}=\dfrac{1}{11}-\dfrac{1}{12},...,\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

So : 

\(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{10}-\dfrac{1}{100}=\dfrac{9}{100}\)

=> S = \(\dfrac{9}{100}\)