Question

Cho A=\(\frac{1}{3^2}+\frac{1}{5^5}+\frac{1}{7^2}+...+\frac{1}{2019^2}\)

Chứng minh A<\(\frac{1}{2}\)

Answer 1

Ta có :

\(A=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)

Cho \(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)

\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)

\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1009}{4040}< \frac{1}{2}\)

Mà A < S ⇒ đpcm