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 :
Now let’s add parentheses using the sequence . Then . That is, we now have the sequence
So the resulting series does converge. However, the original series can’t converge.
The obvious fault is that the terms don’t get smaller. And we know that must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms 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 and , where the first is obtained from the second by removing parentheses using the function . Assume that , and also that there is some so that each of the is a sum of fewer than of the . That is, . 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
We know from last time that , and so if the first series converges then the second one must as well. We need to show that if exists, then we also have .
To this end, pick an . Since the sequence of converge to , we can choose some so that for all . Since the sequence of terms converges to zero, we can increase until we also have for all .
Now take any . Then falls between and for some . We can see that , and that is definitely above . So the partial sum is the sum of all the up through , minus those terms past . That is
But this first sum is just the partial sum , while each term of the second sum is bounded in size by our assumptions above. We check
But since is between and , there must be fewer than terms in this last sum, all of which are bounded by . So we see
and thus we have established the limit.