Question

Chứng minh rằng : 1/2² + 1/3² + 1/4² + ... + 1/100² < 1

Answer 1

Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

        \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)

 

Answer 2

Có: \(\frac{1}{n\left(n+1\right)}< \frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\left(n>1\right)\)

\(\Rightarrow\frac{1}{n}-\frac{1}{n+1}< \frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{100}< A< 1-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow0< A< \frac{99}{100}\)

Có: \(\frac{99}{100}< 1\Rightarrow A< 1\)

Answer 3

\(A=\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{99.100}\\ \)

\(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{99}-\frac{1}{100}\\ =1-\frac{1}{100}\\ =\frac{99}{100}\)

Vì \(\frac{99}{100}< 1\Rightarrow A< 1\)