Ta có: $B = \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}$
$\Leftrightarrow 2B = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}$
$\Leftrightarrow 2B - B = \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}\right) - \left(\dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}\right)$
$\Leftrightarrow B = 1 - \dfrac{1}{2^{99}} < 1$
