Limit of triple sum Suppose one has the following triple sum:
$$S_n=\sum_{s=0}^n\sum_{t=0}^s\sum_{u=0}^sf(n,t)g(n,u)$$
where for all $n$, $-\alpha< S_n <\alpha$ for some real constant $\alpha<\infty$. Since $S_n$ is bounded above and below by a constant may one interchange the limit with the first summand, obtaining
$$\lim_{n\to\infty}S_n=\sum_{s=0}^{\infty}\lim_{n\to\infty}\left(\sum_{t=0}^s\sum_{u=0}^sf(n,t)g(n,u)\right)?$$
Since the limit is now inside the first summand, may one now consider $s$ as a constant and thus bring the limit inside the two other summands to the right of it, yielding
$$\lim_{n\to\infty}S_n=\sum_{s=0}^{\infty}\sum_{t=0}^s\sum_{u=0}^s\left(\lim_{n\to\infty}f(n,t)g(n,u)\right)?$$
If not, why not?
 A: You have $S_n = \sum\limits_{s=0}^n a_{s,n}$ where $a_{s,n}= \sum\limits_{t=0}^s \sum\limits_{u=0}^s f(n,t)g(n,u)$ and you want to 'interchange'
$$\lim\limits_{n\to\infty}S_n = \lim\limits_{n\to\infty}\sum\limits_{s=0}^n \,a_{s,n} \stackrel{?}{=} \sum\limits_{s=0}^{\infty}\,\lim\limits_{n\to\infty}a_{s,n}$$
In principle, there's no reason the third expression makes much sense. In fact, the meaning of $\sum\limits_{s=0}^{\infty}$ is precisely $\lim\limits_{k\to\infty}\sum\limits_{s=0}^k$, so the final expression may be written as:
$$\sum\limits_{s=0}^{\infty}\,\lim\limits_{n\to\infty}a_{s,n}=\lim\limits_{k\to\infty}\left(\sum\limits_{s=0}^k\lim\limits_{n\to\infty}a_{s,n}\right)$$
Now, there are many things missing for us to make sense out of this.


*

*Does $\lim_{n\to\infty}S_n = S_{\infty}$ exist? The $S_n$ may be bounded, but that alone is not sufficient.

*For each $s \in \mathbb{N}$, does $\lim_{n\to\infty}a_{s,n} = a_{s,\infty}$ exist?

*If they all do, does $\lim_{k\to\infty}\sum_{s=0}^ka_{s,\infty} = \stackrel{\displaystyle\sim}{S_{\infty}}$ exist?

*Moreover, if they exist, do the limits coincide, that is, $S_{\infty}=\stackrel{\displaystyle\sim}{S_{\infty}}\,$?
A rough idea of the things that may be happening here... If the first bullet is true, then:
$$\forall \epsilon > 0,\, \exists n'\in \mathbb{N}, \, \forall n \geq n', \, \left\lvert \sum\limits_{s=0}^na_{s,n} - S_{\infty}\right\rvert < \epsilon$$
If the second bullet is true, then for each $s \in \mathbb{N}$ we have that:
$$\forall \epsilon > 0,\, \exists n_s\in \mathbb{N}, \, \forall n \geq n_s, \, \left\lvert a_{s,n} - a_{s,\infty}\right\rvert < \epsilon$$
Finally, if the third bullet is true, then:
$$\forall \epsilon > 0,\, \exists \stackrel{\displaystyle\sim}{n}\in \mathbb{N}, \, \forall n \geq \stackrel{\displaystyle\sim}{n}, \, \left\lvert \sum\limits_{s=0}^na_{s,\infty} - \stackrel{\displaystyle\sim}{S_{\infty}}\right\rvert < \epsilon$$
Assuming all that, were we to try and prove the equality in the fourth bullet, that is, were we to estimate $\left\lvert S_{\infty}-\stackrel{\displaystyle\sim}{S_{\infty}}\right\rvert$, do we have the tools to show that given some $\epsilon>0$ we can take some $n \geq n',\stackrel{\displaystyle\sim}{n}, n_0,n_1,n_2,\dots$? Couldn't it be that this sequence is unbounded?
What I'm trying to show here is that, right now, the problem is a little too loosely posed to perhaps yield any meaningful answer, especially without any motivation, context or background.
