Tham khảo
` S=\frac{1}{16}+\frac{1}{36}+...+\frac{1}{(2n)^2}`
`⇒S=\frac{1}{4}(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2})`
Áp dụng `\frac{1}{n^2}<\frac{1}{n(n+1)}`
`⇒S<\frac{1}{4}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})`
Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}`
`⇒S<\frac{1}{4}.(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1})`
`⇒S<\frac{1}{4}.(1-\frac{1}{n+1})`
Vì `1-\frac{1}{n+1}<2`
`⇒S<\frac{1}{4}.2=\frac{1}{2}`
`\text{©CBT}`
