Question

Bài 20 : Chứng minh rằng 

C = 1/5² + 1/6² + 1/7² +.....+ 1/2008² < 1/4

Answer 1

Ta có

 \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)

.....

\(\frac{1}{2008^2}< \frac{1}{2007.2008}=\frac{1}{2007}-\frac{1}{2008}\)

\(\Rightarrow C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2008^2}< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2008}\)

\(\Rightarrow C< \frac{1}{4}-\frac{1}{2008}< \frac{1}{4}\)

Vậy \(C< \frac{1}{4}\)