Question

Bài 19 : Chứng minh rằng :

C = 1/5^2 + 1/6^2 + 1/7^2 + ....+ 1/2008^2 <1/4

Answer 1

\(C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2008^2}\)

\(< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2007.2008}\)

\(=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{2008-2007}{2007.2008}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(=\frac{1}{4}-\frac{1}{2007}< \frac{1}{4}\).