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

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 $\dfrac{99}{100}<1$
$\Rightarrow A<1$