Chứng minh 1/2^2 + 1/3^2 +…+ 1/100^2 < 1

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1. Giả sử: $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{100^2}$

   Ta có: 

   $\dfrac{1}{2^2}<\dfrac{1}{1.2}$

   …….

   $\dfrac{1}{100^2}<\dfrac{1}{99.100}$

   $⇒A<\dfrac{1}{1.2}+….+\dfrac{1}{99.100}$

   $⇒A<1-\dfrac{1}{2}+….+\dfrac{1}{99}-\dfrac{1}{100}$

   $⇒A<1-\dfrac{1}{100}$

   $⇒A<\dfrac{99}{100}$

   mà $\dfrac{99}{100}<1$

   $⇒A<1$

   Vậy $\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{100^2}<1$

    

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2. Đáp án: $\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}<1$

   Giải thích các bước giải:

   Đặt $A=\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+…+\dfrac{1}{100^{2}}$
   Ta có: 
   $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$
   $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$
   $\dfrac{1}{4^{2}}<\dfrac{1}{3.4}$
   …
   $\dfrac{1}{100^{2}}<\dfrac{1}{99.100}$
   $\Rightarrow A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}$
   $\Leftrightarrow A<1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{99}-\dfrac{1}{100}$
   $\Leftrightarrow A<1-\dfrac{1}{100}$
   $\Leftrightarrow A<\dfrac{99}{100}$
   mà $\dfrac{99}{100}<1$
   $\Rightarrow A<1$

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