# 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?

## 3 Answers

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.