Double Infinite Series

$\begin{aligned} S & = \sum_{x=0}^\infty \sum_{y=0}^\infty \frac {2^{-x-y}}{1+x+y} \\ & = \underbrace{\sum_{x=0}^\infty \frac {2^{-x}}{1+x}}_{y=0} + \underbrace{\sum_{\color{#3D99F6}x=0}^\infty \frac {2^{-x-1}}{2+x}}_{y=1} + \underbrace{\sum_{\color{#3D99F6}x=0}^\infty \frac {2^{-x-2}}{3+x}}_{y=2} + \cdots \\ & = \sum_{x=0}^\infty \frac {2^{-x}}{1+x} + \sum_{\color{#D61F06}x=1}^\infty \frac {2^{-x}}{1+x} + \sum_{\color{#D61F06}x=2}^\infty \frac {2^{-x}}{1+x} + \cdots \\ & = \sum_{x=0}^\infty \frac {(x+1) 2^{-x}}{1+x} \\ & = \sum_{x=0}^\infty 2^{-x} \\ & = \frac 1{1-\frac 12} \\ & = \boxed{2} \end{aligned}$

Let the sum be $\displaystyle S=\sum_{x,y \ge 0} \dfrac{2^{-(x+y)}}{1+x+y}$ , substituting $x+y=k$ we have :

$\displaystyle S= \sum_{k=0}^\infty 2^{-k} =2$ and thus the answer.