# The Unapologetic Mathematician

## Associativity in Series I

As we’ve said before, the real numbers are a topological field. The fact that it’s a field means, among other things, that it comes equipped with an associative notion of addition. That is, for any finite sum we can change the order in which we perform the additions (though not the order of the terms themselves — that’s commutativity).

The topology of the real numbers means we can set up sums of longer and longer sequences of terms and talk sensibly about whether these sums — these series — converge or not. Unfortunately, this topological concept ends up breaking the algebraic structure in some cases. We no longer have the same freedom to change the order of summations.

When we write down a series, we’re implicitly including parentheses all the way to the left. Consider the partial sums:

$\displaystyle s_n=\sum\limits_{k=0}^na_k=((...(((a_0+a_1)+a_2)+a_3)...+a_{n-1})+a_n)$

But what if we wanted to add up the terms in a different order? Say we want to write

$\displaystyle s_6=(((a_0+a_1)+(a_2+a_3))+((a_4+a_5)+a_6))$

Well this is still a left-parenthesized expression, it’s just that the terms are not the ones we looked at before. If we write $b_0=a_0+a_1$, $b_1=a_2+a_3$, and $b_2=a_4+a_5+a_6$ then we have

$\displaystyle s_6=((b_0+b_1)+b_2)=\sum\limits_{j=0}^2b_j=t_2$

So this is actually a partial sum of a different (though related) series whose terms are finite sums of terms from the first series.

More specifically, let’s choose a sequence of stopping points: an increasing sequence of natural numbers $d(j)$. In the example above we have $d(0)=1$, $d(1)=3$, and $d(3)=6$. Now we can define a new sequence

$\displaystyle b_0=\sum\limits_{k=0}^{d(0)}a_k$
$\displaystyle b_j=\sum\limits_{k=d(j-1)+1}^{d(j)}a_k$

Then the sequence of partial sums $t_m$ of this series is a subsequence of the $s_n$. Specifically

$\displaystyle t_m=\sum\limits_{j=0}^mb_j=\sum\limits_{k=0}^{d(0)}a_k+\sum\limits_{j=1}^m\left(\sum\limits_{k=d(j-1)+1}^{d(j)}a_k\right)=\sum\limits_{k=0}^{d(m)}a_k=s_{d(m)}$

We say that the sequence $t_m$ is obtained from the sequence $s_n$ by “adding parentheses” (most clearly notable in the above expression for $t_m$). Alternately, we say that $s_n$ is obtained from $t_m$ by “removing parentheses”.

If the sequence $s_n$ converges, so must the subsequence $t_m=s_{d(m)}$, and moreover to the same limit. That is, if the series $\sum_{k=0}^\infty a_k$ converges to $s$, then any series $\sum_{j=0}^\infty b_j$ obtained by adding parentheses also converges to $s$.

However, convergence of a subsequence doesn’t imply convergence of the sequence. For example, consider $a_k=(-1)^k$ and use $d(j)=2j+1$. Then $s_n$ jumps back and forth between zero and one, but $t_m$ is identically zero. So just because a series converges, another one obtained by removing parentheses may not converge.

May 6, 2008 - Posted by | Analysis, Calculus