Ta có:

\(\frac{1}{2.2}\)<\(\frac{1}{1.2}\)

\(\frac{1}{3.3}\)<\(\frac{1}{2.3}\)

..............

\(\frac{1}{1009.1009}\)<\(\frac{1}{1008.1009}\)

=>A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)

=\(\frac{1}{1}-\frac{1}{1009}=\frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)

=>A<\(\frac{3}{4}\)

Mình nghĩ bạn cần xem lại :

\(A< \frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)không có nghĩa là \(A< \frac{3}{4}\)

Xem lại ..

Ta có : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}< 1\)(đpcm)

+)Ta thấy:\(\frac{1}{2.2}< \frac{1}{1.2}\)

                   \(\frac{1}{3.3}< \frac{1}{2.3}\)

                     ............................

                     ..............................

                  \(\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{99}-\frac{1}{100}< 1\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+.............+\frac{1}{100.100}< 1\left(\text{Đ}PCM\right)\)

Chúc bạn học tốt

rút gọn

Ta có: \(n.n!=\left(n+1-1\right).n!=\left(n+1\right)!-n!\)

Suy ra \(A=\left(2!-1!\right)+\left(3!-2!\right)+...+\left(100!-99!\right)\)\(=100!-1!\)

Vậy.... (chắc hết rút gọn được rồi nhỉ)

=2.1!-1!+3.2!-2!+4.3!-3!+...+101.100!-100!

=2!-1!+3!-2!+4!-3!+...+101!-100!

=101!-1

1.1!+2.2!+3.3!+4.4!+5.5!+2.1.2+3.1.2.3+4.1.2.3.4+5.1.2.3.4.5

=1+4+18+96+600=600+96+4+18+1=600+100+19+=719