The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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May 6, 2008 - Posted by | Analysis, Calculus

2 Comments »

  1. [...] in Series II We’ve seen that associativity doesn’t hold for infinite sums the way it does for finite sums. We can always “add parentheses” to a convergent [...]

    Pingback by Associativity in Series II « The Unapologetic Mathematician | May 7, 2008 | Reply

  2. [...] 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 [...]

    Pingback by Commutativity in Series I « The Unapologetic Mathematician | May 8, 2008 | Reply


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