Ta có: \(3^2>2\cdot4\Rightarrow\frac{1}{3^2}< \frac{1}{2\cdot4}\)
\(5^2>4\cdot6\Rightarrow\frac{1}{5^2}< \frac{1}{4\cdot6}\)
...
\(n^2>n^2-1=\left(n-1\right)\left(n+1\right)\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}\)
Vậy,
\(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2011^2}< \frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2010\cdot2012}\)
\(=\frac{4-2}{2\cdot4}+\frac{6-4}{4\cdot6}+\frac{8-6}{6\cdot8}+...+\frac{2012-2010}{2010\cdot2012}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2010}-\frac{1}{2012}=\frac{1}{2}-\frac{1}{2012}=\frac{1006-1}{2012}=\frac{1005}{2012}\)
_ĐPCM
