Tham khảo
Đặt `A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}`
`⇒A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}`
Áp dụng `\frac{1}{n^2}<\frac{1}{n(n-1)}`
`⇒A<\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}`
Áp dụng `\frac{1}{n(n-1)}=\frac{1}{n}-\frac{1}{n-1}`
`⇒A<\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}`
`⇒A<\frac{1}{4}+\frac{1}{2}-\frac{1}{100}`
`⇒A<\frac{3}{4}-\frac{1}{100}`
`⇒A<\frac{3}{4}`
`\text{©CBT}`
