\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< 1-\frac{1}{100}< 1\)
=> đpcm
Đặt A=1/22 + 1/32 + 1/42 ......+ 1/1002
B=1/1.2+1/2.3+...+1/99.100
Ta có:A=1/22 + 1/32 + 1/42 ......+ 1/1002<B=1/1.2+1/2.3+...+1/99.100 (1)
Mà B=1/1.2+1/2.3+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (2)
Từ (1) và (2) =>A<B<1
=>A<1 (Đpcm)
