What does this double summation with mod evaluate to? Let $X = \{1,2,3,\dots\}$ and $Y = \{0,1\}$.
Define $f:X\times Y \rightarrow \mathbb{R}$ by
$$
f(x,y) = 
\begin{cases}
-2^{-x} &\text{ if } \mod(x, 2) = y \\
(y+1) 2^{-x} &\text{ if } \mod(x,2) = y+1 \\
0 &\text{ otherwise} 
\end{cases}
$$
I've been able to calculate that 
$$
\sum_{x}\sum_{y}f(x,y) = -\frac{1}{3}
$$
How do I compute the opposite summation? Namely 
$$
\sum_{y}\sum_{x}f(x,y)
$$

Note you have
$$\begin{equation}\begin{aligned}
\sum_{x}f(x,0) & = 2^{-1} - 2^{-2} + 2^{-3} - 2^{-4} + \ldots \\
& = \frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{2}\left(\right)^{i} \\
& = \frac{\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)} \\
& = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} \\
& = \frac{1}{3}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned}
\sum_{x}f(x,1) & = -2^{-1} + 2^{-1} - 2^{-3} + 2^{-3} + \ldots \\
& = (-2^{-1} + 2^{-1}) + (-2^{-3} + 2^{-3}) + \ldots \\
& = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note I was able to do the bracketing above due to the series being absolutely convergent (as the sum of the absolute values would be that of a geometric series with $r = \frac{1}{4}$, so its sum would be $\frac{4}{3}$), as explained in the Rearrangements and unconditional convergence section of Wikipedia's "absolute convergence" article. Thus,
$$\begin{equation}\begin{aligned}
\sum_{y}\sum_{x}f(x,y) & = \sum_{y}\left(\sum_{x}f(x,y)\right) \\
& = \sum_{x}f(x,0) + \sum_{x}f(x,1) \\
& = \frac{1}{3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
