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Suppose we have $a_{i,k}\in[0,\infty)$ for $i,k=1,2,3,\dots$. Let's arbitrarily arrange those into a sequence $b_{n}$ for $n=1,2,3,\dots$. I know that the sum $\sum_{n=1}^\infty b_{n}$ is the same no matter how we arranged $a_{i,k}$ into $b_{n}$ (or if the sum is finite or not). Is it true that $\sum_{n=1}^\infty b_{n}=\sum_{i=1}^\infty \sum_{k=1}^\infty a_{i,k}$? If so, why?

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  • $\begingroup$ They’re the same because every number that appears on the left is also on the right. $\endgroup$ – Clayton Mar 22 '19 at 19:16
  • $\begingroup$ Take $S_n$ the double sum up to $n$ in each index. We get an increasing sequence of positive numbers with a limit finite or infinite being its supremum. Then show that if $T_n$ is the partial sum of the $b_k$, so again $T_n$ increasing to its supremum, $T_n \leq S_m$ for high $m$, and $S_n \leq T_m$ for high $m$ and conclude $\endgroup$ – Conrad Mar 22 '19 at 20:15
  • $\begingroup$ @Clayton Your argument would not work for conditionally convergent series. $\endgroup$ – user Mar 22 '19 at 22:14
  • $\begingroup$ @user: of course not. It works for nonnegative terms, as the OP points out. $\endgroup$ – Clayton Mar 22 '19 at 22:33
  • $\begingroup$ Got it, thank you. $\endgroup$ – snak Mar 24 '19 at 6:44

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