\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}\)+...+\(\frac{9999}{10000}\)
= (1-\(\frac{1}{4}\)) +(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{10000}\))
= 99 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)) => 99 - A
Dễ thấy A>0 =>S < 99 (1)
Lại có A= \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)
=> A<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...\(\frac{1}{99}\)-\(\frac{1}{100}\)
=>A<1-\(\frac{1}{100}\)<1
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