Giải thích các bước giải:
Ta có : $A = \dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}$
$\to 2A =1+ \dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2014}}+\dfrac{1}{2^{2015}}$
$\to 2A - A = \bigg( 1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{2014}}+\dfrac{1}{2^{2015}}\bigg)-\bigg( \dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\bigg)$
$\to A = 1-\dfrac{1}{2^{2016} }< 1$
Vậy $A<1$