Giải thích các bước giải:
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\> \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2016.2017}+\dfrac{1}{2017.2018}\\= \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}-\dfrac{1}{2018}\\= \dfrac{1}{2}-\dfrac{1}{2018}> \dfrac{1}{2} - \dfrac{1}{10}=\dfrac{2}{5}(1)\\ A = \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}+\dfrac{1}{2016.2017}\\= 1 - \dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2016}-\dfrac{1}{2017}\\=1-\dfrac{1}{2017}<1(2)$
Từ (1) và (2) `=> 2/5 < A < 1 (đpcm)`
