đặt \(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
.............
\(\frac{1}{100^2}<\frac{1}{99.100}\)
Cộng liên tiếp các BĐT trên ta được:
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)
do đó \(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}\right)<1+1=2\left(đpcm\right)\)
\(1 + {1\over 2.2} + {1\over 3.3} +...+ {1\over 100.100} = 1 + {1\over 2.2} + {1\over 3.3} +...+ {1\over 100.100}\)
\(< 1 + {1\over 1.2} + {1\over 2.3} + {1\over 3.4} +...+ {1\over 99.100} = 1 + 1 - {1\over 2} + {1\over 2} - {1\over 3} + {1\over 3} - {1\over 4} +...+ {1\over 99} - {1\over 100} = 1 + 1 - {1\over 100} = 2 - {1\over 100} < 2\)(đpcm)