tại sao 1/2 - 1/81 > 1/2
Chuẩn chuẩn. :)
\(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{160^2}=\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{80^2}\right)\)
+) \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{80^2}>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{80}.\frac{1}{81}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{80}-\frac{1}{81}\right)\)
\(=\frac{1}{4}+\frac{1}{3}-\frac{1}{81}>\frac{1}{4}+\frac{1}{3}-\frac{1}{12}=\frac{1}{2}\)
=> \(A=\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{80^2}\right)>\frac{1}{4}.\frac{1}{2}=\frac{1}{8}\)
+) \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{80^2}< \frac{1}{4}+\left(\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{80.79}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{79}-\frac{1}{80}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{80}< \frac{3}{4}\)
=> \(A=\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{80^2}\right)< \frac{1}{4}.\frac{3}{4}=\frac{3}{16}\)
