Rudin 8.3 justification for swapping double summations? In theorem 8.3 in Rudin PMA, Rudin swaps an infinite sum with a finite sum without comment. What is the justification for this step?
$$\lim\limits_{n\to\infty}\sum_{i=1}^{\infty}\sum_{j=1}^n a_{ij}=\lim\limits_{n\to\infty}\sum_{j=1}^n\sum_{i=1}^{\infty} a_{ij}$$




 A: It's probably easier to see if you unpack the finite sum:
$$\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij} = \sum_{i=1}^{\infty}(a_{i1} + a_{i2} + \ldots + a_{in})$$
I'm sure that Rudin has proved by this point that if $\sum_{i=1}^{\infty}c_i$ and $\sum_{i=1}^{\infty}d_i$ converge, then so does $\sum_{i=1}^{\infty}(c_i + d_i)$, and we have
$$\sum_{i=1}^{\infty}(c_i + d_i) + \sum_{i=1}^{\infty}c_i + \sum_{i=1}^{\infty}d_i$$
And since this holds for two sequences $(c_i)$ and $(d_i)$, by induction it holds for $n$ sequences $(a_{i1}), (a_{i2}), \ldots, (a_{in})$. Therefore,
$$\sum_{i=1}^{\infty}(a_{i1} + a_{i2} + \ldots + a_{in}) = \sum_{i=1}^{\infty}a_{i1} + \sum_{i=1}^{\infty}a_{i2} + \ldots + \sum_{i=1}^{\infty}a_{in}$$
Converting this back to the more compact notation:
$$\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij} = \sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{ij}$$
Since this equality holds for every $n$, and we know the LHS has a limit as $n \to \infty$, we can take limits of both sides to conclude that
$$\lim_{n \to \infty}\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij} = \lim_{n \to \infty}\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{ij}$$
A: If in a double summation there involves a finite sum, then the order of the summation can be swapped.
Indeed,
$$\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij}=\sum_{i=1}^{\infty}{(a_{i1}+\ldots+a_{in})}=\sum_{i=1}^{\infty}a_{i1}+\ldots+\sum_{i=1}^{\infty}a_{in}=\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{ij}.$$
