đặt A=1/2^2+1/3^2+1/4^2+...+1/100^2
B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (2)
từ (1),(2)=>A<1
Theo đề bài , ta thấy:
BIỂU THỨC TRÊN BÉ HƠN !
Ta có
1/2^2<1/1.2
1/3^2<1/2.3
...
1/100^2<1/99.100
=>1/2^2+1/3^2+...+1/100^2<1/1.2+1/2.3+...+1/99.100=1-1/100<1(đpcm)
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}+\frac{1}{99.100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\)< \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\) < 1- \(\frac{1}{100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\) < \(\frac{99}{100}<1\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\) < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\) < 1 ( điều phải chứng minh)
1/22 + 1/32 + ...+ 1/1002<1
= 1/2.2 + 1/ 3.3 +...+1/100.100
<1/1.2 + 1/2.3 + ...+1/99.100
<1-1/2 + 1-1/2.3 + ...+1-1/99.100
< 1-1/99.100
<1-1/9900<1
Vậy 1/22 + 1/32 + ....+ 1/1002<1
