The Unapologetic Mathematician

Mathematics for the interested outsider

Commutativity in Series I

We’ve seen that associativity may or may not hold for infinite sums, but it can be improved with extra assumptions. As it happens, commutativity breaks down as well, though the story is a bit clearer here.

First we should be clear about what we’re doing. When we add up a finite list of real numbers, we can reorder the list in many ways. In fact, reorderings of n numbers form the symmetric group S_n. If we look back at our group theory, we see that we can write any element in this group as a product of transpositions which swap neighboring entries in the list. Thus since the sum of two numbers is invariant under such a swap — a+b=b+a — we can then rearrange any finite list of numbers and get the same sum every time.

Now we’re not concerned about finite sums, but about infinite sums. As such, we consider all possible rearrangements — bijections p:\mathbb{N}\rightarrow\mathbb{N} — which make up the “infinity symmetric group S_\infty. Now we might not be able to effect every rearrangement by a finite number of transpositions, and commutativity might break down.

If we have a series with terms a_k and a bijection p, then we say that the series with terms b_k=a_{p(k)} is a rearrangement of the first series. If, on the other hand, p is merely injective, then we say that the new series is a subseries of the first one.

Now, if \sum_{k=0}^\infty a_k is only conditionally convergent, I say that we can rearrange the series to give any value we want! In fact, given x\leq y (where these could also be \pm\infty) there will be a rearrangement b_k=a_{p(k)} so that

\displaystyle\liminf\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=x
\displaystyle\limsup\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=y

First we throw away any zero terms in the series, since those won’t affect questions of convergence, or the value of the series if it does converge. Then let p_n be the nth positive term in the sequence a_k, and let -q_n be the nth negative term.

The two series with positive terms \sum_{k=0}^\infty p_k and \sum_{k=0}^\infty q_k both diverge. Indeed, if one converged but the other did not, then the original series \sum_{k=0}^\infty a_k would diverge. On the other hand, if they both converged then the original series would converge absolutely. Conditional convergence happens when the subseries of positive terms and the subseries of negative terms just manage to balance each other out.

Now we take two sequences x_n and y_n converging to x and y respectively. Since the series of positive terms diverges, they’ll eventually exceed any positive number. We can take just enough of them (say k_1 so that

\displaystyle\sum_{k=0}^{k_1}p_k>y_1

Similarly, we can then take just enough negative terms so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l<x_1

Now take just enough of the remaining positive terms so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k>y_2

and enough negatives so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k-\sum\limits_{l=l_1+1}^{l_2}q_l<x_2

and so on and so forth. This gives us a rearrangement of the terms of the series.

Each time we add positive terms we come within p_{k_j} of y_j, and each time we add negative terms we come within q_{l_j} of x_j. But since the original sequence a_n must be converging to zero (otherwise the series couldn’t converge), so must the p_{k_j} and q_{l_j} be converging to zero. And the sequences x_j and y_j are converging to x and y.

It’s straightforward from here to show that the limits superior and inferior of the partial sums of the rearranged series are as we claim. In particular, we can set them both equal to the same number and get that number as the sum of the rearranged series. So for conditionally convergent series, the commutativity property falls apart most drastically.

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May 8, 2008 - Posted by | Analysis, Calculus

4 Comments »

  1. [...] 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 [...]

    Pingback by Commutativity in Series II « The Unapologetic Mathematician | May 9, 2008 | Reply

  2. [...] of all, if a series converges absolutely, then so does any subseries , where is an injective (but not necessarily bijective!) function from the natural numbers to [...]

    Pingback by Commutativity in Series III « The Unapologetic Mathematician | May 12, 2008 | Reply

  3. [...] if this double 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

  4. [...] I’m going to sort of wave my hands here, motivating it by the fact that absolute convergence makes things [...]

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


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