Commutativity in Series I « The Unapologetic Mathematician

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 $n$ th positive term in the sequence $a_k$ , and let $-q_n$ be the $n$ th 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.

May 8, 2008 - Posted by John Armstrong | Analysis, Calculus | | 2 Comments

2 Comments »

[...] 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
[...] 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

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