Chứng minh: 1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ +$\frac{1}{4²}$ + … + $\frac{1}{100²}$ < 2

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1. 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

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2. 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.

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