Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

\(\Rightarrow A< 1\)

ta có:

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

B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{99}\)-1/100

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

Vì a<b mà B lại bé hơn 1 =>A<1

Do 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/99.100 (1)

Mà 1/2 + 1/3 + 1/4 +...+ 1/100= 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100

= 1-1/100 = 99/100 <1  (2)

Từ 1 và 2 => 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1

hay A < 1

\(VP=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(VP=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)

\(VP=\frac{2}{2}-\frac{1}{2}+\frac{3}{3}-\frac{1}{3}+\frac{4}{4}-\frac{1}{4}+...+\frac{100}{100}-\frac{1}{100}\)

\(VP=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)

\(VP=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=VT\) ( đpcm ) 

Mk nghĩ \(VT=100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\) bn xem lại đề có nhầm ko 

Chúc bạn học tốt ~ 

ko mk thấy đúng mà

ko nhầm đề đâu

giải tương tự như câu hôm qua mình giải

để chứng minh A < \(\frac{1}{10}\). Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(\Rightarrow A^2< \left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(=\frac{1.\left(3.5...99\right)}{2.4.6...100}.\frac{2.4.6...100}{\left(3.5.7...99\right).101}\)

\(=\frac{1}{101}< \frac{1}{10}\)

\(\Rightarrow A^2< \frac{1}{101}< \frac{1}{100}=\frac{1}{10^2}\Rightarrow A< \frac{1}{10}\)

để chứng minh A > \(\frac{1}{15}\). Ta thấy \(A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A^2>\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)

\(=\frac{1.\left(3.5...99\right)}{\left(2.4.6...98\right).100}.\frac{1.\left(2.4...98\right)}{2.\left(3.5...99\right)}\)

\(=\frac{1}{100}.\frac{1}{2}=\frac{1}{200}\)

\(\Rightarrow A^2>\frac{1}{200}>\frac{1}{225}=\frac{1}{15^2}\Rightarrow A>\frac{1}{15}\)

a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)

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

=\(1-\frac{1}{100!}< 1\)

\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)

b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)

=\(2-\frac{1}{99}-\frac{1}{100}< 2\)

\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)