# change of order in summation

I'm reading Rudin PMA Theorem 8.3, and I have one question in mind. In the process of the proof, Rudin says

How can we justify $$\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij} = \sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{ij}$$?

Is it possible to change the order of summation if only one of the summations is infinite?

• Yes that is always possible assuming all the series are convergent. Commented Jan 10, 2021 at 17:56

You have to assume that for each $$j\in\{1,\dots, n\}$$, the following sum exists $$\sum\limits_{i=1}^{\infty}a_{ij}$$. Then, for every $$k\in\Bbb{N}$$, we have (if you want to be really strict this follows by induction on the associativity and commutativity of addition) \begin{align} \sum_{i=1}^k\sum_{j=1}^na_{ij}=\sum_{j=1}^n\sum_{i=1}^ka_{ij}. \end{align} On the RHS, by assumption, for each $$j\in\{1,\dots, n\}$$, the limit $$\lim\limits_{k\to\infty}\sum_{i=1}^ka_{ij}$$ exists (afterall that was my first sentence). Thus, we can now use the fact that "limit of sum is sum of limits" (again strictly speaking that theorem is valid for sum of two sequences, so you would need induction for the case of $$n$$ summands) to get that \begin{align} \lim_{k\to\infty}\sum_{i=1}^k\sum_{j=1}^na_{ij} &=\sum_{j=1}^n\lim_{k\to\infty}\sum_{i=1}^ka_{ij}. \end{align} Or in other words, \begin{align} \sum_{i=1}^{\infty}\sum_{j=1}^na_{ij}&=\sum_{j=1}^n\sum_{i=1}^{\infty}a_{ij}. \end{align}