# Why is $\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}a_{ij})=0$?

In a chapter on double series I am currently reading, at some point the notion of double sums not always being interchangeable is shown by the following example:

Consider $a_{ij}=\delta_{ij}-\delta_{i+1,j}$ for $i,j\in\mathbb{N}$ where $\delta$ is the Kronecker-delta. Then observe that $$\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}a_{ij})=0\neq1=\sum_{j=1}^{\infty}(\sum_{i=1}^{\infty}a_{ij})$$

But I don't see why the first half of this equation is like that and why the shouldn't just be equal. If I write it out, I get that $$\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}a_{ij})=\sum_{i=1}^{\infty}(\delta_{i1}-\delta_{i+1,1}+\delta_{i2}-\delta_{i+1,2}+\delta_{i3}+...+\delta_{i\infty}-\delta_{i+1,\infty})$$ And then by adding the values for $i$ it would seem that all terms cancel except for $\delta_{11}$, since there can be no term to cancel it for you would need $i=0$ for that. Since $\delta_{11}=1$ I don't see how this sum can result in zero. Which term in the sum should cancel the $\delta_{11}$?

Moreover, right hereafter my chapter states a theorem that says that if $\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}|a_{ij}|)<\infty$, then $$\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}|a_{ij}|)=\sum_{j=1}^{\infty}(\sum_{i=1}^{\infty}|a_{ij}|)$$ (for any $a_{ij}$). Clearly the sum with the Kronecker-deltas converges to something less than infinity, but still the theorem doesn't hold for them. So how is that possible?

It helps to imagine the terms of the sum laid out in two dimensions:

1  -1   0   0   0  ...
0   1  -1   0   0  ...
0   0   1  -1   0  ...
0   0   0   1  -1  ...
0   0   0   0   1  ...
:   :   :   :   :  ·.


If you sum each (infinitely wide) row first, each of them is $0$, and the sum of those rows is, of course, $0$.

If you sum each (infinitely tall) column first, the sum of each column is $0$ except for the first column, which sums to $1$. So the sum of the sums of the colums is $1$.

The theorem you describe says that you can switch the order of summation if your double sum is absolutely convergent -- that is, if it converges if you replace every term with its absolute value. The $a_{ij}$ example then becomes

1   1   0   0   0  ...
0   1   1   0   0  ...
0   0   1   1   0  ...
0   0   0   1   1  ...
0   0   0   0   1  ...
:   :   :   :   :  ·.


where every row or column (except for the first column) sums to $2$, and neither the sum of sum or rows nor the sum of sum of columns converge. So the theorem does not apply to this situation.

$a_{ij} = +1$ if $j=i$, $-1$ if $j=i+1$, $0$ otherwise. So for every $i \ge 1$, $\sum_{j=1}^\infty a_{ij} = 0$: you have one term ($j=i$) of $+1$ cancelling one term ($j=i+1$) of $-1$, everything else $0$. Sum that over $i$, and you still have $0$.

On the other hand, $\sum_{i=1}^\infty a_{ij} = 0$ except for the case $j=1$, where there is no $-1$ to cancel the $+1$. So $\sum_{j=1}^\infty \sum_{i=1}^\infty a_{ij} = 1$.

Robert Israel and Henning Makholm's explanation are spot on for this problem. I just want to answer the last bit about the theorem.

Note that for each $j$, there is at least $1$ $a_{ij}$ that is nonzero. Namely, $a_{jj} = 1 - 0 = 1$. This means $\sum_{i=1}^\infty |a_{ij}| \ge 1$ for all $j$.

Thus $\sum_{j=1}^\infty \sum_{i=1}^\infty |a_{ij}| \ge \sum_{j=1}^\infty 1 = \infty$. So the theorem you referenced does not apply. The absolute values prevent cancellation between terms.