A Certain Double Sum In this problem, let $a_{i,j}$ be defined piecewise: $a_{i,j} = 0$ if $i <j$; $a_{i,j} = -1$ if $i=j$; and $a_{i,j} = 2^{j-i}$ if $i>j$.
I have to find: $A_1 = \sum_{i=1}^\infty \sum_{j=1}^\infty a_{i,j}$ and $A_2 = \sum_{j=1}^\infty \sum_{i=1}^\infty a_{i,j}$. They should be different.
But I don't know how to handle double series, especially if I have not proven convergence (which I think is the key point).
How does one calculate these? I have never been taught and have no idea.
This is my best guess:
$A_1 = \sum_{i=1}^\infty (-1 + \sum_{j=2}^\infty a_{i,j}) = \sum_{i=1}^\infty (-1+0) = - \infty$.
And likewise for $A_2$. But when I do this, I am getting that both series diverge to $-\infty$.
I know that they should be different. So, what is going wrong? How does one handle these?
 A: In order to visualize both sums, see the following diagram:
$$\begin{array}{cccccccc}
\downarrow&-1&0&0&0&\cdots&\longrightarrow&\mbox{Sum}\downarrow\\\
\downarrow&2^{-1}&-1&0&0&\cdots&\longrightarrow&\\
\downarrow&2^{-2}&2^{-1}&-1&0&\cdots&\longrightarrow&\\
\vdots&2^{-3}&2^{-2}&2^{-1}&-1&\cdots&\longrightarrow&\\
\mbox{Sum}\rightarrow&&
\end{array}$$
Thus $A_1$ is sum each row and then sum all these sums:
$\begin{eqnarray}A_1&=&-1+(-1+\frac{1}{2})+(-1+\frac{1}{2}+\frac{1}{2^2})+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3})+...\\
&=&-1-\frac{1}{2}+(-1+\frac{1}{2}(1+\frac{1}{2}))+(-1+\frac{1}{2}(1+\frac{1}{2}+\frac{1}{2^2}))+...\\
&=&-1-\frac{1}{2}+(-1+1-\frac{1}{2^2})+(-1+1-\frac{1}{2^3})+...\\
&=&-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...\\
&=&-2
\end{eqnarray}$
On the other hand, $A_2$ is sum each column and then sum all these sums:
$\begin{eqnarray}
A_2&=&(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+...\\
&=&(-1+1)+(-1+1)+(-1+1)+...\\
&=&0
\end{eqnarray}$
Thus $A_1\neq A_2$.
