Giải thích các bước giải:
Ta có:
$\dfrac{2}{n^2}<\dfrac{2}{n^2-1}=\dfrac{2}{(n-1)(n+1)}=\dfrac{2n}{(n-1)n(n+1)}$
$\to \dfrac{2}{n^2}<\dfrac{(n+1)+(n-1)}{(n-1)n(n+1)}$
$\to \dfrac{2}{n^2}<\dfrac{1}{(n-1)n}+\dfrac{1}{n(n+1)}$
$\to \dfrac{2}{n^2}<\dfrac{n-(n-1)}{(n-1)n}+\dfrac{(n+1)-n}{n(n+1)}$
$\to \dfrac{2}{n^2}<\dfrac{1}{n-1}-\dfrac1n+\dfrac1n-\dfrac{1}{n+1}$
$\to \dfrac{2}{n^2}<\dfrac{1}{n-1}-\dfrac{1}{n+1}$
Áp dụng ta được:
$\dfrac{2}{3^2}<\dfrac{1}{2}-\dfrac14$
$\dfrac{2}{5^2}<\dfrac{1}{4}-\dfrac16$
$\dfrac{2}{7^2}<\dfrac{1}{6}-\dfrac18$
$\dfrac{2}{9^2}<\dfrac{1}{8}-\dfrac1{10}$
...
$\dfrac{2}{2011^2}<\dfrac{1}{2010}-\dfrac1{2012}$
Cộng vế với vế
$\to A<\dfrac12-\dfrac1{2012}$
$\to A<\dfrac{1005}{2012}$
