The Unapologetic Mathematician

Mathematics for the interested outsider

Commutativity in Series II

We’ve seen that commutativity fails for conditionally convergent series. It turns out, though, that things are much nicer for absolutely convergent series. Any rearrangement of an absolutely convergent series is again absolutely convergent, and to the same limit.

Let \sum_{k=0}^\infty a_k be an absolutely convergent series, and let p:\mathbb{N}\rightarrow\mathbb{N} be a bijection. Define the rearrangement b_k=a_{p(k)}.

Now given an \epsilon>0, absolute convergence tells us we can pick an N so that any tail of the series of absolute values past that point is small. That is, for any n\geq N we have


Now for 0\leq n\leq N, the function p^{-1} takes only a finite number of values (the inverse function exists because p is a bijection). Let M be the largest such value. Thus if m>M we will know that p(m)>N. Then for any such m we have


and we know that the sum on the right is finite by the assumption of absolute convergence. Thus the tail of the series of b_j — and thus the series itself — must converge. Now a similar argument to the one we used when we talked about associativity for absolutely convergent series shows that the rearranged series has the same sum as the original.

This is well and good, but it still misses something. We can’t handle reorderings that break up the order structure. For example, we might ask to add up all the odd terms, and then all the even terms. There is no bijection p that handles this situation. And yet we can still make it work.

Unfortunately, I arrive in Maryland having left my references back in New Orleans. For now, I’ll simply assert that for absolutely convergent series we can perform these more general rearrangements, though I’ll patch this sometime.

May 9, 2008 Posted by | Analysis, Calculus | 3 Comments