Interchanging infinite and finite sum, positive terms

Question

My textbook claims that on the extended reals and for a nonnegative infinite double sequence $(a_{n, m})$ we have \begin{align*} \sum_{n = 1}^{\infty} \sum_{m = 1}^{n} a_{n, m} = \sum_{m = 1}^{\infty} \sum_{n = m}^{\infty} a_{n, m} .\end{align*} As a proof it refers to a theorem that states \begin{align*} \sum_{( n, m) \in \mathbb{N}^{2}}^{}a_{n, m} = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty}a_{n, m} \end{align*} Could someone explain how the statement follows from this theorem?

Answer 1

This is one of those "Fubini-type" index manipulations which you'll have lots of opportunities to practice. If you're dealing only with non-negative terms, then summation behaves very nicely; you can interchange any sums/integrals without concern, because everything is either equal and finite, or everything is $\infty$. One way of figuring out how to do the index manipulation is to draw a figure. So, draw a 2-dimensional grid of the $(n,m)$-plane ($n$ horizontally and $m$ vertically). At a point $(n,m)$ on this grid you have a number $a_{n,m}$. If your goal is to do the summation $\sum_{n=1}^{\infty}\sum_{m=1}^na_{n,m}$, then it means you're adding up all the numbers on a certain triangle (draw this out to see what I'm talking about; this is one of those things where a picture is worth a thousand words).

Anyway, if you don't like pictures, then you can resort to a slight extension of the stated theorem (usually it is stated with an extra equality $\sum_{(n,m)}a_{n,m}=\sum_n\sum_ma_{n,m}=\sum_m\sum_na_{n,m}$), by applying it to a specifically chosen doubly-indexed sequence of numbers. Define \begin{align} \alpha_{n,m}&:= \begin{cases} a_{n,m}&\text{if $1\leq m\leq n$}\\ 0&\text{else} \end{cases} \end{align} So, \begin{align} \sum_{n=1}^{\infty}\sum_{m=1}^na_{n,m}&=\sum_{n=1}^{\infty}\left(\sum_{m=1}^na_{n,m}+\sum_{m=n+1}^{\infty}0\right)\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\alpha_{n,m}\\ &=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\alpha_{n,m}\tag{Fubini}\\ &=\sum_{m=1}^{\infty}\left(\sum_{n=1}^{m-1}\alpha_{n,m}+\sum_{n=m}^{\infty}\alpha_{n,m}\right)\\ &=\sum_{m=1}^{\infty}\left(0+\sum_{n=m}^{\infty}a_{n,m}\right)\\ &=\sum_{m=1}^{\infty}\sum_{n=m}^{\infty}a_{n,m}. \end{align} But again, this is the sort of manipulation which becomes obvious after a little practice, since we know only those terms with $m\leq n$ will contribute to the sum.