Đặt: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\)
Ta có: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow A< \frac{1}{4}.\frac{99}{50}\)
\(\Rightarrow A< \frac{99}{200}< \frac{1}{2}\)
Vậy: \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\left(đpcm\right)\)
