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Let [itex]z : \mathbb{N}^2 \to \mathbb{C}[/itex]. Suppose that:

1) for all natural n, [itex]\sum _{j \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

2) for all natural j, [itex]\sum _{n \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

3) [itex]\sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely.

Can we conclude that

4) [itex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely as well, with

[tex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right ) = \sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/tex]

1) for all natural n, [itex]\sum _{j \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

2) for all natural j, [itex]\sum _{n \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

3) [itex]\sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely.

Can we conclude that

4) [itex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely as well, with

[tex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right ) = \sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/tex]

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