In this article, we will investigate under which assumptions, we may re-arrange alaments within a series and under which circumstances, this is forbidden. If a re-arrangement is allowed, we call such a series unconditionally convergent. We will follow a step-by-step approach starting from finite sums. As mentioned in the last article, absolute convergence will be crucial for re-arrangements of real-valued series.
For finite sums, a re-arrangement is always allowed, since the summation is commutative and every re-arrangement can be written as a finite amount of commutations. As an example, consider the sum
If we re-arrange it such that there alternately aa positive and two negative elements, we get
Mathematically, every re-arrangement of this sum can be expressed by a bijection
. The bijection for the above re-arrangement is given by
therefore
So for any finite sum
with any
we can formulate a generalized commutative law:
Math for Non-Geeks: Template:Aufgabe
Series are sums of infinitely many elements, so there might be "infinitely many re-arrangement steps" necessary. And this may cause some trouble! At first, we need to precisely define what we mean by re-arranging those infinitely many elements:
Definition (re-arrangement of a series)
Let
be a bijection. Then, the series
is called a re-arrangement of the series
.
Sowe have a re-arrangement whenever elements of both series can be assigned one-to-one by a bijection (as it is also the case for finite sums).
Example (re-arrangement of a series)
Consider the harmonic series
. A re-arrangement of it could for instance be
which is given by the bijection
It would be nice to have a generalized commutative law also for series. However, re-arrangements might change the limit! Examples are not too easy to find. One of the easiest is the alternating harmonic series
This series converges, as shown in the article "Alternating series test" . Its limit is given by
.
We use the re-arrangement:
Question: What is the bijection
which causes this re-arrangement?
It is
Or, in a more compact form, we have for all
that
Within the article "Computation rules for series " it is shown taht the limit does not change if we set brackets. Hence, we can re-write the series as
we re-formulate a bit more in order to end up with another alternating harmonic series:
So under this re-arrangement, the limit of the series has halved from
to
.
The proof above is only formal. Mathematically, it runs as follows:
Let
be the sequence of partial sums of the alternating harmonic series and
the sequence of partial sums of the re-arranged series. then,
and
are related via
Since
converges to
, the subsequence
also converges to
. For convergent series, we can compute
However, this does not directly imply convergence of the series
to
. The convergence is proven by taking some
. Then there is an
such that
The elements of the re-arranged sequence are denoted by
, so
. In addition,
is a null sequence. Hence, for every
there is an
with
Now, we set
, and obtain for all
that
since for
, there is 
, there is 
, there is 
This establishes the claim
.
Math for Non-Geeks: Template:Aufgabe
Warning
Limits may change if the elements of a series are re-arranged!
And it even gets worse: converging series can be made divergent by re-arrangement:
Consider the following re-arrangement of the alternating harmonic series:
For
there is
So the partial sums of the re-arranged series up to the summand
can be bounded. How many positive and negative elements are there up to this one? Leaving out the first two of the, we get
positive elements. And there are
negative elements. So in the original series, we have to sum up to element number
. For the partial sum
there is hence
So the partial sum is unbounded! For every
there is some
with
. The re-arranged series diverges to
.
So be careful:
Warning
Re-arranging a converging series may make it diverge!
The examples above treated alternating series, where one was able to "put together
" in a way that any real number or even
could be reached. We can circumvent this problem by only allowing positive series elements. But is just allowing positive elements sufficient to avoid any problems which may lead to different limits under re-arrangement? The answer is indeed: yes! We will formulate a theorem about this and prove it.
However, first we consider an example: the series
converges to
. What if we re-arrange the elements in the same way as we did it for the alternating harmonic series? Let's try it out:
Convergence of
is shown by proving that the sequence of partial sums
stays bounded - see Bounded series and convergence. That means, there is a
, such that
for all
. If we could not show that the sequence of partial sums of the re-arranged series
is bounded, we would be done: this sequence is monotonically increasing and if it is bounded, it has to converge. If we can show taht for all
there is an
with
, we have
for all
and hence, boundedness.
And indeed, this can be shown: the first
elements of the re-arranged series can be found within the original series at positions
. This set has a maximum. So if we set
, then
contains all elements in
plus some additional non-negative ones. Therefore,
. In our example, for
, there is
. We have
with maximum
, so there must be
(the 15 elemeents in
contain all 11 elements in
.
Since
is bounded from above by
, so is
and the re-arranged series converges.
Now, are both limits identical? We denote the limit of the original series by
and that of the re-arranged series by
. Since
, the re-arranged series must have a smaller or equal limit to tzhe original one (
). But now, the original series
can also be seen as a re-arrangement of the re-arranged series
. The "back re-arrangement" map
is a bijection from
to
. We copy the argumentation above and get
, where any
is given and a corresponding
can be found. Taking the limit yields
. Since also
, there is
i.e. both limits coincide.
This argumentation can be generalized to any non-negative series:
Theorem (re-arrangement theorem for non-negative series)
Let
be a converging series with
for all
. Then any re-arrangement converges to the same limit.
Proof (re-arrangement theorem for non-negative series)
Proof step: The re-arranged series converges:
Since
converges, the sequence of partial sums
Is bounded. Now, take any re-arrangement series
and denote by
its partial sums. We set
and get
for all
. Hence, the sequence of partial sums
is bounded, monotonically increasing and we have convergence.
Proof step: The re-arranged series has the same limit:
Hint
If all elements are non-positive, the same argumentation can be applied and we get identical limits for al re-arrangements.
Based on its behaviour under re-arrangement, we define
Definition (unconditional and conditional convergence of a series)
A converging series is called unconditionally convergent if any re-arrangement of it converges to the same limit. Conversely, a series where there is a re-arrangement with a different limit (or which diverges) is called conditionally convergent.
Example (unconditional and conditional convergence of a series)
The alternating harmonic series
is conditionally convergent. The series
is unconditionally convergent.
So the re-arrangement theorem above can equivalently be formulated as follows:
What if there are positive and negative elements within a series? When can we be sure that any re-arrangement yields the same result? The answer is: if and only if it is absolutely convergent. An example is the series
. The corresponding series of absolute values is
and converges. Since every absolutely convergent series converges, the series
converges, as well.
Now, we are interested in proving that its limit is invariant under re-arrangement. In the article Absolute convergence of a series, we proved that a series
is absolutely convergent if and only if it can be split into converging series of non-negative elements
and non-positive elements
. Now,
converges absolutely, so the series
and
converge, as well. As both are purely non-negative or non-positive, we can re-arrange them without changing the limit:
and
If we put both parts together, we obtain
So the entire series can also be re-arranged without changing the limit.
There is
and
So
and
now, the above argumentation holds for any absolutely convergent series, so we can use it to prove a theorem:
Theorem (Re-arrangement theorem for absolutely convergent series)
Let
be an absolutely convergent series. Then, any re-arrangement converges to the same limit.
Proof (Re-arrangement theorem for absolutely convergent series)
Proof step: The re-arranged series converges:
Proof step: The re-arranged series has the same limit:
As shown within the article Absolute convergence of a series, a series
converges absolutely, if and only if its non-negative part
and its non-positive part
converge. By step 1, any re-arrangement of
converges, so
and
converge, as well. As they are non-negative/ non-positive, their limit is invariant under re-arrangement:
and
So
That means, the re-arranged series has the same limit as the original one.
Re-arranging convergent, but not absolutely convergent series
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So, if there is absolute convergence, then the limit of a series is invariant under re-arrangement. Can the limit also be invariant under re-arrangement if the series does not converge absolutely? The answer to this question is actually no! Absolute convergence is equivalent to the limit being invariant under re-arrangement. Even further:
Why does this hold? A series
being not absolutely convergent is equivalent to
or
from above being divergent, see the article Absolute convergence of a series. We even have:
Theorem
If
is a series converging, but not converging absolutely, then the series
and
are both divergent.
Proof
Proof by contraposition: Assume that
or
converges.
Case 1:
and
converge.
Then,
converges absolutely, which cannot be the case.
Case 2:
converges and
diverges.
Then also
diverges. But now, if
and
converge, then also
would have to converge, which is excluded in this case.
This theorem can be used to show that for any convergent, but non-divergent series, we can construct a diverging re-arrangement. The idea it to use the "infinite budgets"
and
and combine them in a way that one wins over the other. For instance, consider our "favourite example": the alternating harmonic series
. We construct a re-arrangement
, which diverges to
the following way: We sum up a lot of positive terms, until we surpass
. Then a negative term follows. Then, we sum up sufficiently many terms to get above
. A negative summand follows and then again enough summands to surpass
and so on...
The result will diverge to
: after
has been passed, we can get at most down by 1 again and always stay above
. This argument holds for arbitrarily large
and hence yields divergence. For the alternating harmonic series, the re-arrangement looks as follows:
So for any
there is an
with
for all
and we get divergence.
This argument holds for any conditionally convergent series:
Theorem
Let
be a convergent series which does not converge absolutely. Then there is a rte-arrangement of this series which diverges.
Hint
Within the proof, we have even shown that
Interchanging
and
we analogously have that: