Chứng minh:
1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ + … + $\frac{1}{100²}$ < 2
Chứng minh: 1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ + … + $\frac{1}{100²}$ < 2
By Eva
By Eva
Chứng minh:
1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ + … + $\frac{1}{100²}$ < 2
1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ +…+$\frac{1}{100²}$ < 2
Ta có:
1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ +…+$\frac{1}{100²}$
< 1 + $\frac{1}{1.2}$ + $\frac{1}{2.3}$ +$\frac{1}{3.4}$ +…+$\frac{1}{99.100}$
= 1 + 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{4}$ +…+$\frac{1}{99}$ – $\frac{1}{100}$
= 2 – $\frac{1}{100}$ < 2
A < 2 – $\frac{1}{100}$ < 2
⇒ A < 2
Vậy A < 2
@Kimetsu No Yaiba
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.