Let $\{a_{mn}\}$ be a double series, where $a_{mn}>0$ for all $m,n\in\Bbb{N}$. If $\sum\limits_{i=1}^\infty{a_{ik}}$ is finite for all $k\in\Bbb{N}$ and $\sum\limits_{j=1}^\infty{a_{hj}}$ is finite for all $j\in\Bbb{N}$, then $\sum\limits_{i=1}^\infty \sum\limits_{j=1}^\infty{a_{ij}}=\sum\limits_{j=1}^\infty\sum\limits_{i=1}^\infty{a_{ij}}$.
Is the above statement true?
If it is, how does one go about proving it?
Actually, for nonnegative sequence $\{a_{mn}\}$, we always have $$\sum_{n=1}^\infty\sum_{m=1}^\infty a_{mn}=\sum_{m=1}^\infty\sum_{n=1}^\infty a_{mn}.$$
This can be proved by monotone convergence theorem. Note that both sides could be $+\infty$.
For general sequence $\{a_{mn}\}$, if $\sum_{n=1}^\infty\sum_{m=1}^\infty|a_{mn}|<\infty$, then $$\sum_{n=1}^\infty\sum_{m=1}^\infty a_{mn}=\sum_{m=1}^\infty\sum_{n=1}^\infty a_{mn}.$$
This can be proved by Fubini's Theorem or Lebesgue Dominated Convergence Theorem.
If a double series is absolutely convergent, it is also convergent; moreover, any series obtained by rearrangement of its terms also converges, https://www.encyclopediaofmath.org/index.php/Double_series
