Question

1) CMR : \(A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{100^2}< 1\frac{3}{4}\)

Answer 1

\(A=\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{100^2}< \frac{1}{1}+\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\\ \)

\(\frac{1}{1}+\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\\ =\frac{1}{1}+\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\\ =\frac{1}{1}+\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\\ =1\frac{3}{4}-\frac{1}{100}< 1\frac{3}{4}\)

Vậy \(A< 1\frac{3}{4}\)

Answer 2

Ta có với mọi n là số tự nhiên thì : \(\frac{1}{n^2}< \frac{1}{n\left(n+1\right)}\)

Áp dụng : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}< 1< 1\frac{3}{4}\)