Đặt A = $1+\frac{1}{2^2}$ + $\frac{1}{3^2}$ + $\frac{1}{4^2}$ + ... + $\frac{1}{100^2}$
=> A - 1 = $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + $\frac{1}{4^2}$ + ... + $\frac{1}{100^2}$
=> A - 1 < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{99.100}$
=> A - 1 < 1 - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{99}$ - $\frac{1}{100}$
=> A - 1 < 1 - $\frac{1}{100}$
=> A - 1 < $\frac{99}{100}$
=> A < $\frac{199}{100}$ < 2
Vậy A<2.
