Suppose one has the following triple sum:
Sn=n∑s=0s∑t=0s∑u=0f(n,t)g(n,u)
where for all n, −α<Sn<α for some real constant α<∞. Since Sn is bounded above and below by a constant may one interchange the limit with the first summand, obtaining
limn→∞Sn=∞∑s=0limn→∞(s∑t=0s∑u=0f(n,t)g(n,u))?
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
limn→∞Sn=∞∑s=0s∑t=0s∑u=0(limn→∞f(n,t)g(n,u))?
If not, why not?
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