Math for Non-Geeks/Rearrangement theorem for series

{{#invoke:Math for Non-Geeks/Seite|oben}}

In this article, we will investigate under which assumptions, we may re-arrange alaments within a series and under which circumstances, this is forbidden. Template:Noprint 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.

Re-arrangement of finite sums

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 $\sigma :\{1,2,3,4,5,6,7,8\}\to \{1,2,3,4,5,6,7,8\}$. The bijection for the above re-arrangement is given by

therefore

So for any finite sum $\sum _{k=1}^n{a}_k$ with any $n\in \mathbb{N}$ we can formulate a generalized commutative law:

The problem with series

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:

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).

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 $S=\ln (2)$.

We use the re-arrangement:

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 $S$ to $\frac{1}{2}S$.

Re-arranged series might diverge

And it even gets worse: converging series can be made divergent by re-arrangement:

So be careful:

Re-arrangement for non-negative series

The examples above treated alternating series, where one was able to "put together $\mathrm{\infty }-\mathrm{\infty }$" in a way that any real number or even $\pm \mathrm{\infty }$ 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.

Unconditional and conditional convergence

Based on its behaviour under re-arrangement, we define

So the re-arrangement theorem above can equivalently be formulated as follows:

Re-arranging absolutely convergent series

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 $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k^2}$. The corresponding series of absolute values is $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ and converges. Since every absolutely convergent series converges, the series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k^2}$ 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 $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is absolutely convergent if and only if it can be split into converging series of non-negative elements $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{+}$and non-positive elements $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{-}$ . Now, $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{\sigma (k+1)}}{\sigma (k{)}^2}=\sum _{k=1}^{\mathrm{\infty }}{a}_{\sigma (k)}$ converges absolutely, so the series $\sum _{k=1}^{\mathrm{\infty }}{a}_{\sigma (k)}^{+}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_{\sigma (k)}^{-}$ 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.

Re-arranging convergent, but not absolutely convergent series

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 $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ being not absolutely convergent is equivalent to $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{+}$ or $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{-}$ from above being divergent, see the article Absolute convergence of a series. We even have:

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" $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{+}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_k^{-}$ and combine them in a way that one wins over the other. For instance, consider our "favourite example": the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$ . We construct a re-arrangement $\sum _{k=1}^{\mathrm{\infty }}{a}_{\sigma (k)}$, which diverges to $\mathrm{\infty }$ the following way: We sum up a lot of positive terms, until we surpass $n=1$ . Then a negative term follows. Then, we sum up sufficiently many terms to get above $n=2$ . A negative summand follows and then again enough summands to surpass $n=3$ and so on...

The result will diverge to $\mathrm{\infty }$ : after $n\in \mathbb{N}$ has been passed, we can get at most down by 1 again and always stay above $n-1$. This argument holds for arbitrarily large $n-1$ and hence yields divergence. For the alternating harmonic series, the re-arrangement looks as follows:

So for any $(n-1)\in \mathbb{N}$ there is an $M\in \mathbb{N}$ with $\sum _{k=1}^m{a}_{\sigma (k)}\ge n$ for all $m>M$ and we get divergence.

This argument holds for any conditionally convergent series:

{{#invoke:Math for Non-Geeks/Seite|unten}}