A problem by أحمد الحلاق

Convert each term in $A$ and $B$ to:

$A \Rightarrow \log_{1/2^{2^k}} 2 = \frac{\log_{2} 2}{\log_{2} 1/2^{2^k}} = \frac{1}{log_{2} 2^{-2^k}} = -(\frac{1}{2})^{k}$ (i)

$B \Rightarrow 2^{-\log_{2} 2^{k}} = 2^{\log_{2} (1/2^k)} = (\frac{1}{2})^k$ (ii)

such that $A+B = \Sigma_{k=1}^{\infty} (\frac{1}{2})^k - (\frac{1}{2})^k = \boxed{0}.$