# Math for Non-Geeks/Rearrangement theorem for series

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.

## 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
${\begin{aligned}\sum _{k=1}^{8}(-1)^{k+1}{\frac {1}{k}}&={\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{5}}}{\color {blue}-{\frac {1}{6}}}{\color {OliveGreen}+{\frac {1}{7}}}{\color {blue}-{\frac {1}{8}}}={\frac {533}{840}}\end{aligned}}$
If we re-arrange it such that there alternately aa positive and two negative elements, we get
${\begin{aligned}{\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}{\color {OliveGreen}+{\frac {1}{5}}+{\frac {1}{7}}}={\frac {533}{840}}\end{aligned}}$
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
$\sigma (1)=1,\ \sigma (2)=2,\ \sigma (3)=4,\ \sigma (4)=3,\ \sigma (5)=6,\ \sigma (6)=8,\ \sigma (7)=5,\ \sigma (8)=7$
therefore
$\underbrace {\sum _{k=1}^{8}(-1)^{k+1}{\frac {1}{k}}} _{{\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{5}}}{\color {blue}-{\frac {1}{6}}}{\color {OliveGreen}+{\frac {1}{7}}}{\color {blue}-{\frac {1}{8}}}}=\underbrace {\sum _{k=1}^{8}(-1)^{\sigma (k)+1}{\frac {1}{\sigma (k)}}} _{{\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}{\color {OliveGreen}+{\frac {1}{5}}+{\frac {1}{7}}}}$
So for any finite sum $\sum _{k=1}^{n}a_{k}$ with any $n\in \mathbb {N}$ we can formulate a generalized commutative law:
$\sum _{k=1}^{n}a_{k}=\sum _{k=1}^{n}a_{\sigma (k)}$
Math for Non-Geeks: Template:Aufgabe

## 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:
Definition (re-arrangement of a series)
Let $\sigma :\mathbb {N} \to \mathbb {N}$ be a bijection. Then, the series $\sum _{k=1}^{\infty }a_{\sigma (k)}$ is called a re-arrangement of the series $\sum _{k=1}^{\infty }a_{k}$.
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 $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$. A re-arrangement of it could for instance be
${\tfrac {1}{2}}+1+{\tfrac {1}{4}}+{\tfrac {1}{3}}+{\tfrac {1}{6}}+{\tfrac {1}{5}}+{\tfrac {1}{8}}+{\tfrac {1}{7}}+\ldots$
which is given by the bijection
$\sigma (2k-1)=2k{\text{ and }}\sigma (2k)=2k-1{\text{ for }}k\in \mathbb {N}$
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
$\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}={\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{5}}}{\color {blue}-{\frac {1}{6}}}{\color {OliveGreen}+{\frac {1}{7}}}{\color {blue}-{\frac {1}{8}}}{\color {OliveGreen}+{\frac {1}{9}}}{\color {blue}-{\frac {1}{10}}}{\color {OliveGreen}+}\ldots {\color {OliveGreen}+{\frac {1}{2n-1}}}{\color {blue}-{\frac {1}{2n}}}\pm \ldots$
This series converges, as shown in the article "Alternating series test" . Its limit is given by $S=\ln(2)$.
We use the re-arrangement:
${\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {OliveGreen}+{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}{\color {OliveGreen}+}\ldots {\color {OliveGreen}+{\frac {1}{2n-1}}}{\color {blue}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\pm \ldots$
Question: What is the bijection $\sigma :\mathbb {N} \to \mathbb {N}$ which causes this re-arrangement?
It is $\sigma (1)={\color {OliveGreen}1},\ \sigma (2)={\color {blue}2},\ \sigma (3)={\color {blue}4},\ \sigma (4)={\color {OliveGreen}3},\ \sigma (5)={\color {blue}6},\ \sigma (6)={\color {blue}8},\ \sigma (7)={\color {OliveGreen}5},\ \sigma (8)={\color {blue}10},\ \sigma (9)={\color {blue}12},\ldots$
Or, in a more compact form, we have for all $n\in \mathbb {N}$ that
$\sigma (3n-2)={\color {green}2n-1},\ \sigma (3n-1)={\color {blue}4n-2},\ \sigma (3n)={\color {blue}4n}.$
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
$\left({\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{2n-1}}}{\color {blue}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\right)\pm \ldots$
we re-formulate a bit more in order to end up with another alternating harmonic series:
${\begin{aligned}&\left({\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{2n-1}}}{\color {blue}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\right)\pm \ldots \\=&\left({\frac {1}{2}}{\color {blue}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\left({\frac {1}{6}}{\color {blue}-{\frac {1}{8}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}\left({\frac {1}{4n-2}}{\color {blue}-{\frac {1}{4n}}}\right)\pm \ldots \\=&{\frac {1}{2}}\left(1{\color {blue}-{\frac {1}{2}}}\right){\color {OliveGreen}+}{\frac {1}{2}}\left({\frac {1}{3}}{\color {blue}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}{\frac {1}{2}}\left({\frac {1}{2n-1}}{\color {blue}-{\frac {1}{2n}}}\right)\pm \ldots \\=&{\frac {1}{2}}\left(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\ldots +{\frac {1}{2n-1}}-{\frac {1}{2n}}\pm \ldots \right)\\=&{\frac {1}{2}}\left(\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}\right)\end{aligned}}$
So under this re-arrangement, the limit of the series has halved from $S$ to ${\tfrac {1}{2}}S$.
The proof above is only formal. Mathematically, it runs as follows:
Let $(S_{N})_{N\in \mathbb {N} }$ be the sequence of partial sums of the alternating harmonic series and $(T_{N})_{N\in \mathbb {N} }$ the sequence of partial sums of the re-arranged series. then, $T_{3n}$ and $S_{2n}$ are related via ${\begin{aligned}T_{3n}&=\left({\color {OliveGreen}1}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{3}}}{\color {blue}-{\frac {1}{6}}-{\frac {1}{8}}}\right){\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{5}}}{\color {blue}-{\frac {1}{10}}-{\frac {1}{12}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}\left({\color {OliveGreen}{\frac {1}{2n-1}}}{\color {blue}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\right)\\&=\left({\frac {1}{2}}{\color {blue}-{\frac {1}{4}}}\right){\color {OliveGreen}+}\left({\frac {1}{6}}{\color {blue}-{\frac {1}{8}}}\right){\color {OliveGreen}+}\left({\frac {1}{10}}{\color {blue}-{\frac {1}{12}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}\left({\frac {1}{4n-2}}{\color {blue}-{\frac {1}{4n}}}\right)\\&={\frac {1}{2}}\left(1{\color {blue}-{\frac {1}{2}}}\right){\color {OliveGreen}+}{\frac {1}{2}}\left({\frac {1}{3}}{\color {blue}-{\frac {1}{4}}}\right){\color {OliveGreen}+}{\frac {1}{2}}\left({\frac {1}{5}}{\color {blue}-{\frac {1}{6}}}\right){\color {OliveGreen}+}\ldots {\color {OliveGreen}+}{\frac {1}{2}}\left({\frac {1}{2n-1}}{\color {blue}-{\frac {1}{2n}}}\right)\\&={\frac {1}{2}}\left(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\ldots +{\frac {1}{2n-1}}-{\frac {1}{2n}}\right)\\&={\frac {1}{2}}\left(\sum _{k=1}^{2n}(-1)^{k+1}{\frac {1}{k}}\right)\\&={\frac {1}{2}}S_{2n}\end{aligned}}$
Since $(S_{N})$ converges to $S$ , the subsequence $(S_{2n})$ also converges to $S$. For convergent series, we can compute
${\frac {1}{2}}S=\lim _{n\to \infty }{\frac {1}{2}}S_{2n}=\lim _{n\to \infty }T_{3n}$
However, this does not directly imply convergence of the series $(T_{N})$ to ${\tfrac {1}{2}}S$ . The convergence is proven by taking some $\epsilon >0$ . Then there is an $N_{1}\in \mathbb {N}$ such that
$|T_{3n}-{\tfrac {1}{2}}S|<\epsilon {\text{ for all }}n\geq N_{1}$
The elements of the re-arranged sequence are denoted by $t_{k}$, so $T_{N}=\sum _{k=1}^{N}t_{k}$. In addition, $(t_{k})_{k\in \mathbb {N} }$ is a null sequence. Hence, for every $\epsilon >0$ there is an $N_{2}\in \mathbb {N}$ with
$|T_{3n+1}-T_{3n}|=|t_{3n+1}|<\epsilon {\text{ and }}|T_{3n+2}-T_{3n}|=|t_{3n+2}+t_{3n+1}|<\epsilon {\text{ for all }}n\geq N_{2}$
Now, we set ${\tilde {N}}=\max\{N_{1},N_{2}\}$, and obtain for all $N\geq 3{\tilde {N}}+2$ that
$|T_{N}-{\tfrac {1}{2}}S|<\epsilon$
since for
* $N=3k$, there is $3{\tilde {N}}+2\geq 3{\tilde {N}}\geq 3N_{1}$
* $N=3k+1$, there is $3{\tilde {N}}+2\geq 3{\tilde {N}}+1\geq 3N_{2}+1$
* $N=3k+2$, there is $3{\tilde {N}}+2\geq 3{\tilde {N}}+2\geq 3N_{2}+2$
This establishes the claim $\lim _{N\to \infty }T_{N}={\tfrac {1}{2}}S$.
Math for Non-Geeks: Template:Aufgabe
Warning
Limits may change if the elements of a series are re-arranged!

## Re-arranged series might diverge

And it even gets worse: converging series can be made divergent by re-arrangement:
Consider the following re-arrangement of the alternating harmonic series:
${\color {blue}1}{\color {red}-{\frac {1}{2}}}{\color {blue}+{\frac {1}{3}}}{\color {red}-{\frac {1}{4}}}{\color {blue}+{\frac {1}{5}}+{\frac {1}{7}}}{\color {red}-{\frac {1}{6}}}{\color {blue}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{15}}}{\color {red}-{\frac {1}{8}}}{\color {blue}+}\ldots {\color {blue}+{\frac {1}{2^{n}+1}}+{\frac {1}{2^{n}+3}}+\ldots +{\frac {1}{2^{n+1}-1}}}{\color {red}-{\frac {1}{2n+2}}}\pm \ldots$
For $n\in \mathbb {N}$ there is
${\color {blue}{\frac {1}{2^{n}+1}}+{\frac {1}{2^{n}+3}}+\ldots +{\frac {1}{2^{n+1}-1}}}\geq \underbrace {{\frac {1}{2^{n+1}}}+{\frac {1}{2^{n+1}}}+\ldots +{\frac {1}{2^{n+1}}}} _{2^{n-1}{\text{ times}}}={\frac {2^{n-1}}{2^{n+1}}}={\frac {1}{4}}\cdot {\frac {2^{n-1}}{2^{n-1}}}={\frac {1}{4}}$
So the partial sums of the re-arranged series up to the summand ${\color {red}-{\tfrac {1}{2n+2}}}$ can be bounded. How many positive and negative elements are there up to this one? Leaving out the first two of the, we get ${1+2+4+\ldots +2^{n-1}=\sum _{k=0}^{n-1}2^{k}}{\underset {\text{sum formula}}{\overset {\text{geometric}}{=}}}{\color {blue}2^{n}-1}$ positive elements. And there are ${\color {red}n}$ negative elements. So in the original series, we have to sum up to element number $2^{n-1}+n+1$ . For the partial sum $(T_{2^{n-1}+n+1})_{n\in \mathbb {N} }$ there is hence
${\begin{aligned}T_{2^{n-1}+n+1}&={\color {blue}1}{\color {red}-{\frac {1}{2}}}{\color {blue}+\underbrace {\frac {1}{3}} _{\geq {\frac {1}{4}}}}{\color {red}-{\frac {1}{4}}}{\color {blue}+\underbrace {{\frac {1}{5}}+{\frac {1}{7}}} _{\geq {\frac {1}{4}}}}{\color {red}-{\frac {1}{6}}}{\color {blue}+\underbrace {{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{15}}} _{\geq {\frac {1}{4}}}}{\color {red}-{\frac {1}{8}}}{\color {blue}+}\ldots {\color {blue}+\underbrace {{\frac {1}{2^{n}+1}}+{\frac {1}{2^{n}+3}}+\ldots +{\frac {1}{2^{n+1}-1}}} _{\geq {\frac {1}{4}}}}{\color {red}-{\frac {1}{2n+1}}}\\&\geq \underbrace {{\color {blue}1}{\color {red}-{\frac {1}{2}}}} _{={\frac {1}{2}}}\underbrace {{\color {blue}+{\frac {1}{4}}}{\color {red}-{\frac {1}{4}}}} _{=0}\underbrace {{\color {blue}+{\frac {1}{4}}}{\color {red}-{\frac {1}{6}}}} _{={\frac {1}{12}}}\underbrace {{\color {blue}+{\frac {1}{4}}}{\color {red}-{\frac {1}{8}}}} _{\geq {\frac {1}{12}}}{\color {blue}+}\ldots \underbrace {{\color {blue}+{\frac {1}{4}}}{\color {red}-{\frac {1}{2n+1}}}} _{\geq {\frac {1}{12}}}\\&\geq {\frac {1}{2}}+(n-1)\cdot {\frac {1}{12}}\end{aligned}}$
So the partial sum is unbounded! For every $C>0$ there is some $m=2^{n-1}+n+1\in \mathbb {N}$ with $S_{m}\geq C$. The re-arranged series diverges to $\infty$.
So be careful:
Warning
Re-arranging a converging series may make it diverge!

## Re-arrangement for non-negative series

The examples above treated alternating series, where one was able to "put together $\infty -\infty$" in a way that any real number or even $\pm \infty$ 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.
However, first we consider an example: the series $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ converges to $\zeta (2)={\frac {\pi ^{2}}{6}}$. 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:
${{\color {OliveGreen}1}{\color {red}+{\frac {1}{2^{2}}}}{\color {OliveGreen}+{\frac {1}{3^{2}}}}{\color {red}+{\frac {1}{4^{2}}}}{\color {OliveGreen}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}}{\color {red}+{\frac {1}{6^{2}}}}{\color {OliveGreen}+{\frac {1}{9^{2}}}+{\frac {1}{11^{2}}}+{\frac {1}{13^{2}}}+{\frac {1}{15^{2}}}}{\color {red}+{\frac {1}{8^{2}}}}{\color {OliveGreen}+}\ldots {\color {OliveGreen}+{\frac {1}{(2^{n}+1)^{2}}}+{\frac {1}{(2^{n}+3)^{2}}}+\ldots +{\frac {1}{(2^{n+1}-1)^{2}}}}{\color {red}+{\frac {1}{(2n+2)^{2}}}}\pm \ldots }$
Convergence of $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ is shown by proving that the sequence of partial sums $(S_{n})_{n\in \mathbb {N} }=\left(\sum _{k=1}^{n}{\tfrac {1}{k^{2}}}\right)_{n\in \mathbb {N} }$ stays bounded - see Bounded series and convergence. That means, there is a $K>0$, such that $S_{n}\leq K$ for all $n\in \mathbb {N}$ . If we could not show that the sequence of partial sums of the re-arranged series $(T_{m})_{m\in \mathbb {N} }=\left(\sum _{k=1}^{m}a_{\sigma (k)}\right)_{m\in \mathbb {N} }$ 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 $m\in \mathbb {N}$ there is an $n\in \mathbb {N}$ with $T_{m}\leq S_{n}$ , we have $T_{M}\leq K$ for all $m\in \mathbb {N}$ and hence, boundedness.
And indeed, this can be shown: the first $m$ elements of the re-arranged series can be found within the original series at positions $\{\sigma (1),\sigma (2),\ldots ,\sigma (m)\}\subset \mathbb {N}$. This set has a maximum. So if we set $n=\max\{\sigma (1),\sigma (2),\ldots ,\sigma (m)\}$, then $S_{n}$ contains all elements in $T_{m}$ plus some additional non-negative ones. Therefore, $T_{m}\leq S_{n}$. In our example, for $m=11$, there is $T_{11}={\color {OliveGreen}1}{\color {red}+{\frac {1}{2^{2}}}}{\color {OliveGreen}+{\frac {1}{3^{2}}}}{\color {red}+{\frac {1}{4^{2}}}}{\color {OliveGreen}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}}{\color {red}+{\frac {1}{6^{2}}}}{\color {OliveGreen}+{\frac {1}{9^{2}}}+{\frac {1}{11^{2}}}+{\frac {1}{13^{2}}}+{\frac {1}{15^{2}}}}$. We have $\{\sigma (1),\sigma (2),\ldots ,\sigma (11)\}=\{1,2,\ldots ,15\}$ with maximum $n=15$, so there must be $T_{11}\leq S_{15}$ (the 15 elemeents in $S_{15}$ contain all 11 elements in $>T_{11}$.
Since $(S_{n})_{n\in \mathbb {N} }$ is bounded from above by $K$, so is $(T_{m})_{m\in \mathbb {N} }$ and the re-arranged series converges.
Now, are both limits identical? We denote the limit of the original series by $\lim _{n\to \infty }S_{n}=S$ and that of the re-arranged series by $\lim _{m\to \infty }T_{m}=S'$. Since $T_{m}\leq S_{n}$, the re-arranged series must have a smaller or equal limit to tzhe original one ($S'\leq S$). But now, the original series $\sum _{k=1}^{\infty }a_{k}$ can also be seen as a re-arrangement of the re-arranged series $\sum _{k=1}^{\infty }a_{\sigma (k)}$ . The "back re-arrangement" map $\mu =\sigma ^{-1}$ is a bijection from $\mathbb {N}$ to $\mathbb {N}$. We copy the argumentation above and get $S_{n}\leq T_{m'}$, where any $n\in \mathbb {N}$ is given and a corresponding $m'\in \mathbb {N}$ can be found. Taking the limit yields $S\leq S'$. Since also $S'\leq S$, there is $S'=S$ i.e. both limits coincide.
This argumentation can be generalized to any non-negative series:
Theorem (re-arrangement theorem for non-negative series)
Let $\sum _{k=1}^{\infty }a_{k}$ be a converging series with $a_{k}\geq 0$ for all $k\in \mathbb {N}$. 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 $\sum _{k=1}^{\infty }a_{k}$ converges, the sequence of partial sums $(S_{n})_{n\in \mathbb {N} }=\left(\sum _{k=1}^{n}a_{k}\right)_{n\in \mathbb {N} }$ Is bounded. Now, take any re-arrangement series $\sum _{k=1}^{\infty }a_{\sigma (k)}$ and denote by $T_{m}$ its partial sums. We set $n=\max\{\sigma (1),\sigma (2),\ldots ,\sigma (m)\}$ and get $T_{m}\leq S_{n}$ for all $m\in \mathbb {N}$. Hence, the sequence of partial sums $(T_{m})_{m\in \mathbb {N} }$ is bounded, monotonically increasing and we have convergence.
Proof step: The re-arranged series has the same limit:
Denote by $S$ the limit of the original series and by $S'$ teh limit of the re-arranged series. By step 1, there is $S'\leq S$. But now, the original series is just a re-arrangement of the re-arranged one (the corresponding bijection is $\mu =\sigma ^{-1}$ with $\sum _{k=1}^{n}a_{k}=\sum _{k=1}^{n}a_{\sigma ^{-1}(\sigma (k))}=\sum _{k=1}^{n}a_{\mu (\sigma (k))}$). So the same argumentation as in step 1 also yields $S\leq S'$ and both limits coincide.
Hint
If all elements are non-positive, the same argumentation can be applied and we get identical limits for al re-arrangements.

### Unconditional and conditional convergence

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 $\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}$ is conditionally convergent. The series $\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}$ is unconditionally convergent.
So the re-arrangement theorem above can equivalently be formulated as follows:
Theorem (Re-arrangement theorem for non-negative series (alternative formulation))
Let $\sum _{k=1}^{\infty }a_{k}$ be a convergent series with $a_{k}\geq 0$ for all $k\in \mathbb {N}$. Then, the series converges unconditionally.

## 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 $\sum _{k=1}^{\infty }{\tfrac {(-1)^{k+1}}{k^{2}}}$. The corresponding series of absolute values is $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ and converges. Since every absolutely convergent series converges, the series $\sum _{k=1}^{\infty }{\tfrac {(-1)^{k+1}}{k^{2}}}$ 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 $\sum _{k=1}^{\infty }a_{k}$ is absolutely convergent if and only if it can be split into converging series of non-negative elements $\sum _{k=1}^{\infty }a_{k}^{+}$and non-positive elements $\sum _{k=1}^{\infty }a_{k}^{-}$ . Now, $\sum _{k=1}^{\infty }{\tfrac {(-1)^{\sigma (k+1)}}{\sigma (k)^{2}}}=\sum _{k=1}^{\infty }a_{\sigma (k)}$ converges absolutely, so the series $\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}$ and $\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}$ converge, as well. As both are purely non-negative or non-positive, we can re-arrange them without changing the limit:
$\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}=\sum _{k=1}^{\infty }a_{k}^{+}$
and
$\sum _{k=1}^{\infty }\underbrace {a_{\sigma (k)}^{-}} _{\leq 0}=-\left(-\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}\right)=-\sum _{k=1}^{\infty }\underbrace {-a_{\sigma (k)}^{-}} _{\geq 0}=-\sum _{k=1}^{\infty }-a_{k}^{-}=\sum _{k=1}^{\infty }a_{k}^{-}$
If we put both parts together, we obtain
${\begin{aligned}\sum _{k=1}^{\infty }a_{\sigma (k)}&=\sum _{k=1}^{\infty }(a_{\sigma (k)}^{+}+a_{\sigma (k)}^{-})=\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}+\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}\\[0.3em]&=\sum _{k=1}^{\infty }a_{k}^{+}+\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{k=1}^{\infty }(a_{k}^{+}+a_{k}^{-})=\sum _{k=1}^{\infty }a_{k}\end{aligned}}$
So the entire series can also be re-arranged without changing the limit.
Question: What are the both parts $\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}$ and $\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}$ of the series above $\sum _{k=1}^{\infty }a_{\sigma (k)}=\sum _{k=1}^{\infty }{\tfrac {(-1)^{\sigma (k+1)}}{\sigma (k)^{2}}}$?
There is
$a_{k}^{+}={\begin{cases}{\frac {1}{k^{2}}}&{\text{ for odd }}k{\text{, i.e. }}k=2n+1\\0&{\text{ for even }}k\end{cases}}$
and
$a_{k}^{-}={\begin{cases}0&{\text{ for odd }}k\\-{\tfrac {1}{k^{2}}}&{\text{ for even }}k{\text{, i.e. }}k=2n\end{cases}}$
So
$\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}=\sum _{k=1}^{\infty }a_{k}^{+}=\sum _{n=1}^{\infty }{\frac {1}{(2n+1)^{2}}}$
and
$\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}=\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{n=1}^{\infty }{\frac {-1}{(2n)^{2}}}$
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 $\sum _{k=1}^{\infty }a_{k}$ 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:
Let $\sum _{k=1}^{\infty }a_{k}$ be absolutely convergent, so $\sum _{k=1}^{\infty }|a_{k}|$converges. Since $|a_{k}|\geq 0$ , the re-arrangement theorem for non-negative series can be applied and any re-arrangement $\sum _{k=1}^{\infty }|a_{\sigma (k)}|$ converges, as well. Since absolutely convergent series are also convergent, the series $\sum _{k=1}^{\infty }a_{\sigma (k)}$ converges, as well.
Proof step: The re-arranged series has the same limit:
As shown within the article Absolute convergence of a series, a series $\sum _{k=1}^{\infty }a_{k}$ converges absolutely, if and only if its non-negative part $\sum _{k=1}^{\infty }a_{k}^{+}$ and its non-positive part $\sum _{k=1}^{\infty }a_{k}^{-}$ converge. By step 1, any re-arrangement of $\sum _{k=1}^{\infty }a_{\sigma (k)}$ converges, so $\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}$ and $\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}$ converge, as well. As they are non-negative/ non-positive, their limit is invariant under re-arrangement:
$\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}=\sum _{k=1}^{\infty }a_{k}^{+}$
and
$\sum _{k=1}^{\infty }\underbrace {a_{\sigma (k)}^{-}} _{\leq 0}=-\left(-\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}\right)=-\sum _{k=1}^{\infty }\underbrace {-a_{\sigma (k)}^{-}} _{\geq 0}=-\sum _{k=1}^{\infty }-a_{k}^{-}=\sum _{k=1}^{\infty }a_{k}^{-}$
So
${\begin{aligned}\sum _{k=1}^{\infty }a_{\sigma (k)}&=\sum _{k=1}^{\infty }(a_{\sigma (k)}^{+}+a_{\sigma (k)}^{-})=\sum _{k=1}^{\infty }a_{\sigma (k)}^{+}+\sum _{k=1}^{\infty }a_{\sigma (k)}^{-}\\[0.3em]&=\sum _{k=1}^{\infty }a_{k}^{+}+\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{k=1}^{\infty }(a_{k}^{+}+a_{k}^{-})=\sum _{k=1}^{\infty }a_{k}\end{aligned}}$
That means, the re-arranged series has the same limit as the original one.

## 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 $\sum _{k=1}^{\infty }a_{k}$ being not absolutely convergent is equivalent to $\sum _{k=1}^{\infty }a_{k}^{+}$ or $\sum _{k=1}^{\infty }a_{k}^{-}$ from above being divergent, see the article Absolute convergence of a series. We even have:
Theorem
If $\sum _{k=1}^{\infty }a_{k}$ is a series converging, but not converging absolutely, then the series $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{-}$ are both divergent.
Proof
Proof by contraposition: Assume that $\sum _{k=1}^{\infty }a_{k}^{+}$ or $\sum _{k=1}^{\infty }a_{k}^{-}$ converges.
Case 1: $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{-}$ converge.
Then, $\sum _{k=1}^{\infty }a_{k}$ converges absolutely, which cannot be the case.
Case 2: $\sum _{k=1}^{\infty }a_{k}^{+}$ converges and $\sum _{k=1}^{\infty }a_{k}^{-}$ diverges.
Then also
$\sum _{k=1}^{\infty }a_{k}^{+}+\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{k=1}^{\infty }a_{k}$
diverges. But now, if $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{+}+\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{k=1}^{\infty }a_{k}$ converge, then also
$\sum _{k=1}^{\infty }a_{k}^{-}=\sum _{k=1}^{\infty }a_{k}-\sum _{k=1}^{\infty }a_{k}^{+}$
would have to converge, which is excluded in this case.
Case 3: $\sum _{k=1}^{\infty }a_{k}^{+}$ diverges and $\sum _{k=1}^{\infty }a_{k}^{-}$ converges
can not occur by the same arguments as in step 2, where the roles of $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{-}$ are exchanged.
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" $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{-}$ and combine them in a way that one wins over the other. For instance, consider our "favourite example": the alternating harmonic series $\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}$ . We construct a re-arrangement $\sum _{k=1}^{\infty }a_{\sigma (k)}$, which diverges to $\infty$ 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 $\infty$ : 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:
${\begin{aligned}{\color {OliveGreen}a_{1}}&{\color {OliveGreen}={\frac {1}{1}}=1=a_{\sigma (1)}}&\Longrightarrow &\sum _{k=1}^{1}a_{\sigma (k)}={\color {OliveGreen}1}\geq {\color {Blue}1}\\{\color {Red}a_{2}}&{\color {Red}=-{\frac {1}{2}}=a_{\sigma (2)}}&\Longrightarrow &\sum _{k=1}^{2}a_{\sigma (k)}={\color {OliveGreen}1}{\color {Red}-{\frac {1}{2}}}={\color {Orange}{\frac {1}{2}}}\\{\color {OliveGreen}a_{3}}&{\color {OliveGreen}={\frac {1}{3}}=a_{\sigma (3)},\ a_{5}={\frac {1}{5}}=a_{\sigma (4)},\ldots ,a_{41}={\frac {1}{41}}=a_{\sigma (22)}}&\Longrightarrow &\sum _{k=1}^{22}a_{\sigma (k)}={\color {OliveGreen}1}{\color {Red}-{\frac {1}{2}}}{\color {OliveGreen}+{\frac {1}{3}}+{\frac {1}{5}}+\ldots {\frac {1}{41}}}={\color {Blue}2{.}005\geq 2}\\a_{4}&=-{\frac {1}{4}}=a_{\sigma (23)}&\Longrightarrow &\sum _{k=1}^{23}a_{\sigma (k)}=2{.}005-{\frac {1}{4}}=1{.}7505\\a_{43}&={\frac {1}{43}}=a_{\sigma (24)},\ a_{45}={\frac {1}{45}}=a_{\sigma (25)},\ldots &\Longrightarrow &\ldots \end{aligned}}$
So for any $(n-1)\in \mathbb {N}$ there is an $M\in \mathbb {N}$ with $\sum _{k=1}^{m}a_{\sigma (k)}\geq n$ for all $m>M$ and we get divergence.
This argument holds for any conditionally convergent series:
Theorem
Let $\sum _{k=1}^{\infty }a_{k}$ be a convergent series which does not converge absolutely. Then there is a rte-arrangement of this series which diverges.
Proof
As $\sum _{k=1}^{\infty }a_{k}$ converges, but does not converge absolutely, the series $\sum _{k=1}^{\infty }a_{k}^{+}$ and $\sum _{k=1}^{\infty }a_{k}^{-}$ must both diverge (this has been shown above).
Since $\sum \limits _{k=1}^{\infty }a_{k}^{+}$ diverges, we have an $m_{1}\in \mathbb {N}$ with
$\sum \limits _{k=1}^{m_{1}}a_{k}^{+}=\sum \limits _{k=1}^{m_{1}}a_{\sigma (k)}\geq 1$
In addition, since $\sum \limits _{k=1}^{\infty }a_{k}^{-}$ diverges, there is an $m_{2}\in \mathbb {N}$ with
$\sum \limits _{k=1}^{m_{1}}a_{k}^{+}+\sum _{k=1}^{m_{2}}a_{k}^{-}=\sum \limits _{k=1}^{m_{2}}a_{\sigma (k)}\leq 1$
Now, there is an $m_{2}\in \mathbb {N}$ with
$\sum \limits _{k=1}^{m_{2}}a_{\sigma (k)}+\sum \limits _{k=1}^{m_{3}}a_{k}^{+}\geq 2$
Continuing iteratively, we get that for each $N\in \mathbb {N}$ there is an $m\in \mathbb {N}$, such that for that re-arranged series $\sum \limits _{k=1}^{\infty }a_{\sigma (k)}$ there is
$\sum \limits _{k=1}^{m}a_{\sigma (k)}\geq N$
So the constructed series $\sum \limits _{k=1}^{\infty }a_{\sigma (k)}$ diverges.
Hint
Within the proof, we have even shown that
Interchanging $\sum \limits _{k=1}^{\infty }a_{k}^{+}$ and $\sum \limits _{k=1}^{\infty }a_{k}^{-}$ we analogously have that: