! là j z

\("!"\)  là giai thừa đó bạn ạ .

\(VD:\)  \(3!=1.2.3=6\)

          \(4!=1.2.3.4=24\)

nhầm tớ lộn sang bài khác sorry

trình bày cách giải giùm với nhé

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\frac{1}{1\cdot2}< \frac{1}{2^2}\)

\(\frac{1}{2\cdot3}< \frac{1}{3^2}\)

\(\frac{1}{3\cdot4}< \frac{1}{4^2}\)

...

\(\frac{1}{99\cdot100}< \frac{1}{100^2}\)

\(\Rightarrow B< A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(B< A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< A=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}>\frac{3}{4}\)

\(\Leftrightarrow B< \frac{3}{4}< A\)

TA CÓ 1/2^2=1/4

1/3^2<1/2.3=1/2-1/3

1/4^2<1/3.4=1/3-1/4

1/100^2<1/99.100

=>1/2^2+2/3^2+.....+1/100^2<1/1.2+1/2.3+..+1/99.100

=1-99/100=99/100<1

Ta thấy:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

Mà: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)=1-\frac{1}{100}<\frac{3}{4}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}<\frac{3}{4}\)