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I really don't get why $$\sum_{j=0}^{\infty} \sum_{k=0}^j a_{k,j-k} =\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} a_{k,j}$$

The only thing that makes sense to me is that if I write the terms I have

$$\sum_{j=0}^{\infty} \sum_{k=0}^j a_{k,j-k} =\sum_{j=0}^{\infty} (a_{0,j}+a_{1,j-1}+...+a_{j-1,1}+a_{j,0})$$

And then rearranging the terms I would eventually get the RHS on the first equation, but I have negative index so I'm not sure if this is rigorous or even correct.

Can someone explain me why the equality holds?

Thanks for your time.

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    $\begingroup$ Without a whiteboard it is hard to explain it, but you are essentially counting dots in the positive quadrant. There are different ways to count it; you can go row by row, or you can count downwards diagonals (which correspond to your two different sums). Also, I believe this equality does not necessarily hold; it will hold if the sequence a is absolutely convergent, or if they are all non-negative. $\endgroup$ – E-A Mar 13 at 1:02
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Summing over $\mathbb Z^2$ diagonally gives the first series while summing vertically gives the second:

\begin{align}S&=\color{purple}{a_{0,0}}+\color{#229955}{a_{0,1}}+\color{purple}{a_{0,2}}+\color{#229955}{a_{0,3}}+\dots\\&\hspace{1pt}+\color{#229955}{a_{1,0}}+\color{purple}{a_{1,1}}+\color{#229955}{a_{1,2}}+\dots\\&\hspace{1pt}+\color{purple}{a_{2,0}}+\color{#229955}{a_{2,1}}+\dots\\&\hspace{1pt}+\color{#229955}{a_{3,0}}+\dots\\&\hspace{1pt}+\dots\\S&=\color{purple}{a_{0,0}}+\color{#229955}{a_{0,1}}+\color{purple}{a_{0,2}}+\color{#229955}{a_{0,3}}+\dots\\&\hspace{1pt}+\color{purple}{a_{1,0}}+\color{#229955}{a_{1,1}}+\color{purple}{a_{1,2}}+\dots\\&\hspace{1pt}+\color{purple}{a_{2,0}}+\color{#229955}{a_{2,1}}+\dots\\&\hspace{1pt}+\color{purple}{a_{3,0}}+\dots\\&\hspace{1pt}+\dots\end{align}

As mentioned in the comments, absolute convergence allows us to sum the infinite series in different orders, but for conditionally converging series they may not be equal.

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