The Unapologetic Mathematician

Mathematics for the interested outsider

Associativity in Series II

I’m leaving for DC soon, and may not have internet access all day. So you get this now!

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 sequence, but we can’t always remove them.

The first example we mentioned last time. Consider the series with terms a_k=(-1)^k:

\displaystyle\sum\limits_{k=0}^\infty a_k=1+(-1)+1+(-1)+...

Now let’s add parentheses using the sequence d(j)=2j+1. Then b_j=(-1)^{(2(j-1)+1)+1}+(-1)^{2j+1}=1+(-1)=0. That is, we now have the sequence

\displaystyle\sum\limits_{j=0}^\infty b_j=(1+(-1))+(1+(-1))+...=0+0+...=0

So the resulting series does converge. However, the original series can’t converge.

The obvious fault is that the terms a_k don’t get smaller. And we know that \lim\limits_{k\rightarrow\infty}a_k must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms b_j go to zero, but when we remove them this condition can fail. And it turns out there’s just one more condition we need so that we can remove parentheses.

So let’s consider the two series with terms a_k and b_j, where the first is obtained from the second by removing parentheses using the function d(j). Assume that \lim_{k\rightarrow\infty}a_k=0, and also that there is some M>0 so that each of the b_j is a sum of fewer than M of the a_k. That is, d(j+1)-d(j)<M. Then the series either both diverge or both converge, and if they converge they have the same sum.

We set up the sequences of partial sums

\displaystyle s_n=\sum\limits_{k=0}^na_k
\displaystyle t_m=\sum\limits_{j=0}^mb_j

We know from last time that t_m=s_{d(m)}, and so if the first series converges then the second one must as well. We need to show that if t=\lim\limits_{m\rightarrow\infty}t_m exists, then we also have \lim\limits_{n\rightarrow\infty}s_n=t.

To this end, pick an \epsilon>0. Since the sequence of t_m converge to t, we can choose some N so that \left|t_m-t\right|<\frac{\epsilon}{2} for all m>N. Since the sequence of terms a_k converges to zero, we can increase N until we also have \left|a_k\right|<\frac{\epsilon}{2M} for all k>N.

Now take any n>d(N). Then n falls between d(m) and d(m+1) for some m. We can see that m\geq N, and that n is definitely above N. So the partial sum s_n is the sum of all the a_k up through k=d(m+1), minus those terms past k=n. That is

\displaystyle s_n=\sum\limits_{k=0}^na_k=\sum\limits_{k=0}^{d(m+1)}a_k-\sum\limits_{n+1}^{d(m+1)}a_k

But this first sum is just the partial sum t_{m+1}, while each term of the second sum is bounded in size by our assumptions above. We check


But since n is between d(m) and d(m+1), there must be fewer than M terms in this last sum, all of which are bounded by \frac{\epsilon}{2M}. So we see


and thus we have established the limit.


May 7, 2008 - Posted by | Analysis, Calculus


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  2. […] series itself — must converge. Now a similar argument to the one we used when we talked about associativity for absolutely convergent series shows that the rearranged series has the same sum as the […]

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  3. […] But this is just the sum of a bunch of absolute values from the original series, and so is bounded by . So the series of absolute values of has bounded partial sums, and so 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. […]

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