Ta có:\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
..........
\(\frac{1}{100^2}<\frac{1}{99.100}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}.............+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}<1\)
Vì \(\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{99.100}\)nên
\(\frac{1}{2^2}+\frac{1}{3^2}+............+\frac{1}{100^2}<1\)
mình quên xin lỗi
Chứng minh rằng :1/2^2+1/3^2+1/4^2+...+1/100^2<1
