Ta có
$A = \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}}$
Suy ra
$\dfrac{1}{2} A = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}}$
Khi đó
$A - \dfrac{1}{2} A = \left( \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}} \right) - \left( \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}} \right)$
Suy ra
$\dfrac{1}{2} A = \dfrac{1}{2} - \dfrac{1}{2^{2017}}$
hay
$A = 1 - \dfrac{1}{2^{2016}}< 1$
Vậy $A < 1$
