Ta có
\(\frac{S}{2}=2+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}\)
\(\Rightarrow\frac{S}{2}=2+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{S}{2}=2+\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{S}{2}=2+\frac{2}{5}=\frac{12}{5}\)
\(\Rightarrow\frac{S}{2}=\frac{6}{5}< 2\)
=> S<2
\(S=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
\(S=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{90}\)
\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\)
\(S=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(S=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(S=2\left(1-\frac{1}{10}\right)\)
có : \(1-\frac{1}{10}< 1\Rightarrow2\left(1-\frac{1}{10}\right)< 2.1\)
Vậy: \(S< 2\)
hư!hu! mấy ông chs bẩn! Nhân lúc t giải bài khác đi giải bài này trước t 
\(\Rightarrow\frac{S}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.6}+..........+\frac{1}{2.45}\)
\(\Rightarrow\frac{S}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...........+\frac{1}{90}\)
\(\Rightarrow\frac{S}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{9.10}\)
\(\Rightarrow\frac{S}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{S}{2}=1-\frac{1}{10}\)
\(\Rightarrow S=\frac{9}{10}.2\)
\(\Rightarrow S=\frac{9}{5}\)
Vì \(\frac{9}{5}< 2\Rightarrow S< 2\)
S = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
= \(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{90}\)
= \(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{2\times4}+\frac{2}{4\times5}+.....+\frac{2}{9\times10}\)
= \(2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{9\times10}\right)\)
= \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2\left(1-\frac{1}{10}\right)\)
Ta có: \(1-\frac{1}{10}< 1=>2\left(1-\frac{1}{10}\right)< 2\times1\)
=> S < 2