The Unapologetic Mathematician

Mathematics for the interested outsider

Commutativity in Series III

Okay, here’s the part I promised I’d finish last Friday. How do we deal with rearrangements that “go to infinity” more than once? That is, we chop up the infinite set of natural numbers into a bunch of other infinite sets, add each of these subseries up, and then add the results up. If the original series was absolutely convergent, we’ll get the same answer.

First of all, if a series \sum_{k=0}^\infty a_k converges absolutely, then so does any subseries \sum_{j=0}^\infty a_{p(j)}, where p is an injective (but not necessarily bijective!) function from the natural numbers to themselves. For instance, we could let p(j)=2j and add up all the even terms from the original series.

To see this, notice that at any finite n we have a maximum value N=\max\limits_{0\leq j\leq n}p(j). Then we find

\displaystyle\left|\sum\limits_{j=0}^na_{p(j)}\right|\leq\sum\limits_{j=0}^n\left|a_{p(j)}\right|\leq\sum\limits_{k=0}^N\left|a_k\right|\leq\sum\limits_{k=0}^\infty\left|a_k\right|

So the new sequence of partial sums of absolute values is increasing and bounded above, and thus converges.

Now let’s let p_0, p_1, p_2, and so on be a countable collection of functions defined on the natural numbers. We ask that

  • Each p_n is injective.
  • The image of p_n is a subset P_k\subseteq\mathbb{N}.
  • The collection \left\{P_0,P_1,P_2,...\right\} is a partition of \mathbb{N}. That is, these subsets are mutually disjoint, and their union is all of \mathbb{N}.

If \sum_{k=0}^\infty a_k is an absolutely convergent series, we define \left(b_n\right)_j=a_{p_n(j)} — the subseries defined by p_n. Then from what we said above, each \sum_{j=0}^\infty\left(b_n\right)_j is an absolutely convergent series whose sum we call s_n. We assert now that \sum_{n=0}^\infty s_n is an absolutely convergent series whose sum is the same as that of \sum_{k=0}^\infty a_k.

Let’s set t_m=\sum_{n=0}^m\left|s_n\right|. That is, we have

\displaystyle t_m\leq\sum\limits_{j=0}^\infty\left|\left(b_1\right)_j\right|+...+\sum\limits_{j=0}^\infty\left|\left(b_m\right)_j\right|=\sum\limits_{j=0}^\infty\left(\left|\left(b_1\right)_j\right|+...+\left|\left(b_m\right)_j\right|\right)

But this is just the sum of a bunch of absolute values from the original series, and so is bounded by \sum_{k=0}^\infty\left|a_k\right|. So the series of absolute values of s_n has bounded partial sums, and so \sum_{n=0}^\infty s_n converges absolutely. That it has the same sum as the original is another argument exactly analogous to (but more complicated than) the one for a simple rearrangement, and for associativity of absolutely convergent series.

This pretty much wraps up all I want to say about calculus for now. I’m going to take a little time to regroup before I dive into linear algebra in more detail than the abstract algebra I covered before. But if you want to get ahead, go back and look over what I said about rings and modules. A lot of that will be revisited and fleshed out in the next sections.

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May 12, 2008 - Posted by | Analysis, Calculus

3 Comments »

  1. [...] series converges absolutely, we can adjust the order of summations freely. Indeed, we’ve seen examples of other rearrangements that all go through as soon as the convergence is [...]

    Pingback by Translating Power Series « The Unapologetic Mathematician | September 16, 2008 | Reply

  2. [...] Like when we translated power series, I’m going to sort of wave my hands here, motivating it by the fact that absolute convergence makes things nice. [...]

    Pingback by Products of Power Series « The Unapologetic Mathematician | September 22, 2008 | Reply

  3. [...] union is finite, but absolute convergence will give us all sorts of flexibility to reassociate and rearrange our [...]

    Pingback by Signed Measures and Sequences « The Unapologetic Mathematician | June 23, 2010 | Reply


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