# Double infinite summation and changing order of summation in series of nonnegative numbers

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?

• They’re the same because every number that appears on the left is also on the right. – Clayton Mar 22 '19 at 19:16
• 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 – Conrad Mar 22 '19 at 20:15
• @Clayton Your argument would not work for conditionally convergent series. – user Mar 22 '19 at 22:14
• @user: of course not. It works for nonnegative terms, as the OP points out. – Clayton Mar 22 '19 at 22:33
• Got it, thank you. – snak Mar 24 '19 at 6:44