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a: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\)
\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac12+\frac13+\cdots+\frac{1}{100}-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\)
\(=1+\frac12+\cdots+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)
b: Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)
Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)
TA có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)
Do đó: \(\frac{1}{76}+\frac{1}{77}+\ldots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)
Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)
=>\(A<\frac56\) (3)
Ta có: \(\frac{1}{51}>\frac{1}{75};\frac{1}{52}>\frac{1}{75};\ldots;\frac{1}{75}=\frac{1}{75}\)
Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (4)
Ta có: \(\frac{1}{76}>\frac{1}{100};\frac{1}{77}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)
Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}=\frac{25}{100}=\frac14\) (5)
Từ (4),(5) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)>\frac13+\frac14\)
=>A>7/12(6)
Từ (3),(6) suy ra 5/6<A<7/12
Xét vế trái: A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=>\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
=>đpcm (VP là vế phải)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+............+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+..........+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+.........+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+.....+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+.........+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(\Rightarrowđpcm\)