Series Rearragment

$\textbf{Theorem}$ Let $\sum a_n$ be a series of real numbers which converges, but not absolutely. If $-\infty \leq \alpha \leq \beta \leq \infty$,

then there exists a rearrangement $\sum a_n^{\prime}$ with partial sums $s_n^{\prime}$ such that $\liminf _{n \rightarrow \infty}s_n^{\prime}=\alpha, \quad \limsup _{n \rightarrow \infty} s_n^{\prime}=\beta .$

Proof. First let's consider any arbitrary sequences $\{c_n\}$ and $\{d_n\}$ with the following: $c_n \ge 0$ and $d_n \ge 0$

and $c_n - d_n =a_n$ and $c_n +d_n = |a_n|$. Then we can claim that $\sum c_n$ and $\sum d_n$ are both divergent.

It's because if $\sum c_n$ and $\sum d_n$ converges, then $\sum c_n+d_n = \sum |a_n|$ converges,

which contradicts to $\sum a_n$ be a series of real numbers which does not converges absolutely;

if $\sum c_n$ diverges and $\sum d_n$ converges, then $\sum a_n +\sum d_n = \sum a_n + d_n = \sum c_n$, which means $c_n$ converges,

contradicting to the assumption; if $\sum c_n$ converges and $\sum d_n$ diverges, we will still have the same thing.

Now, let us define $C_n$ to be zero if $a_n \le 0$ and $C_n$ to be $a_n$ if $a_n >0$ and define $D_n$ to be zero if $a_n \ge 0$ and $D_n$ to be $-a_n$ if $a_n <0$. Then $C_n - D_n$ will be $a_n$

and $C_n +D_n$ will be the absolute value of $a_n$. So $C_n$ diverges and also $D_n$ diverges.

Let's further define $g_n$ to be the smallest i where $C_i >0$ and $i > g_{n-1}$ and $h_n$ to be be the smallest $i$ where $D_i>0$ and $i > h_{n-1}$ and $g_{-1}=h_{-1}=-1.$

Then $C_{g_i}$ and $D_{h_i}$ will be positive real numbers excluding the zero terms from $C_n$ and $D_n$ and also their sum diverges. Since both are positive, we will have

$C_{g_i}$ will go arbitrary large and $D_{h_i}$ will go arbitrary large.

Now, let's consider two sequences $\alpha_n$ goes to $\alpha$ and $\beta_n$ goes to $\beta$. We also assume that $\beta_1>0$ and $\alpha_n < \beta_{n+1}$ for arbitrary $n$. We can construct the following alternating summation.

Define $f(n), g(n)$ as the following with integer $n \ge 1$, in a

$$f(n) = \begin{cases} x & n \text{ is odd} \text{ and } g(n-1) + \sum_{i=f(n-2)}^{x} C_{g_i} > \beta_{\frac{n+1}{2}} \text{ and } x \text{ is the smallest} \\ x & n \text{ is even} \text{ and } g(n-1) - \sum_{i=f(n-2)}^{x} D_{h_i} < \alpha_{\frac{n}{2}} \text{ and } x \text{ is the smallest} \end{cases}$$

$$g(n) = \begin{cases} g(n-1) + \sum_{i=f(n-2)}^{f(n)} C_{g_i} & n \text{ is odd} \\ g(n-1) - \sum_{i=f(n-2)}^{f(n)} D_{h_i} & n \text{ is even} \end{cases}$$

Then we will claim the sequence of $g(n)$ forms a sequence where $$\liminf_{n\to \infty} g(n) = \alpha \text{ and} \limsup_{n\to \infty} g(n) = \beta.$$

When $n$ is odd, we will have $g(n) = g(n-1) + \sum_{i=f(n-2)}^{f(n)} C_{g_i} > \beta_{(n+1)/2}$ and $g(n-1)+\sum_{i=f(n-2)}^{f(n)-1} C_{g_i}$ will either be $g(n-1)$ which is less than $\alpha_{\frac{n-1}{2}} = \alpha_{\frac{n+1}{2}-1} < \beta_{\frac{n+1}{2}}$, or

$g(n-1) +\sum_{i=f(n-2)}^{f(n)-1} C_{g_i} $ which is less than or equal to $\beta_{\frac{n+1}{2}}$, which is by definition. So $0<g(n) - \beta_{\frac{n+1}{2}} =g(n-1) +\sum_{i=f(n-2)}^{f(n)-1}C_{g_i} - \beta_{\frac{n+1}{2}}+ C_{g_i} \le C_{g_{f(n)}},$ which means $|g(n) - \beta_{\frac{n+1}{2}}|\le C_{g_{f(n)}}.$

When $n$ is even, we will have $g(n)= g(n-1) - \sum_{i=f(n-2)}^{f(n)} D_{h_i} < \alpha_{n/2}$ and $g(n-1)-\sum_{i=f(n-2)}^{f(n)-1}D_{h_i} \ge \alpha_{n/2}$. And so $0>g(n) - \alpha_{n/2}=g(n-1)-\sum_{i=f(n-2)}^{f(n)-1}D_{h_i} -a_{n/2}-D_{h_{f(n)}} \ge- D_{h_{f(n)}}$, which means $|g(n) - \alpha_{n/2}| \le D_{h_{f(n)}}$

By sending $n$ to infinity, we will have $C_{g_{f(n)}} \to 0$ and $D_{h_{f(n)}} \to 0$, which are sub-terms from $\{a_n\}_n$ since we require $\sum a_n$ converges. And so there's a subsequence of $\{g(n)\}$, such that $g(n) \to \beta$ and there's a subsequence of $\{g(n)\}$

such that $g(n) \to \alpha.$

To show that there's no subsequence that has a limit that's either less than $\alpha$ or greater than $\beta$. We can showcase one example of no sub-sequential limit less than $\alpha$ and the proof that no limit is greater than or equal to $\beta$ can be implemented in a similar way.

Assume $\gamma < \alpha$ is a subsequence limit of $g(n)$. Since it's a subsequence limit, that means there is a function $f(n): \mathbb{N} \to \mathbb{N}$, such that for every $\epsilon>0$, we will have $\mathcal N$ such for every $n> \mathcal N$, $|g(f(n)) - \gamma|< \epsilon$. Since we have when $n$ is odd, we will have $g(n) > \beta_{\frac{n+1}{2}}$, which is going to $\beta$ and greater than $a_{\frac{n+1}{2}-1},$ we can assert that there's large enough $M$ such that $n>M$, then $f(n)$ are all even with small enough $\epsilon$, but we know that when $n$ is even, we have a unique limit that is $\alpha$. Therefore, we will have a contradiction.

$\textbf{Theorem}$ If $\sum a_n$ is a series of complex numbers which converges absolutely, then every rearrangement of $\sum a_n$ converges, and they all converge to the same sum.

Proof. Since $a_n$ converges absolutely, we will have for every $\epsilon >0$, there's an $N$ such that for every $m>n>N, \sum_{i=n}^m |a_i| < \epsilon.$ In fact, for any $g: \mathbb{N} \leftrightarrow \mathbb{N}_{\ge N+1}$, we will have $\sum_{i=1}^{k} |a_i| <\epsilon$ for any $k$.

Then, for a rearrangement of the series, we will have the rearrangement function as $f(n) : \mathbb{N} \leftrightarrow \mathbb{N}$

we take $N' = 1+\max_{1\le i \le N}f^{-1}(i)$, then we will have for every $n>m >N'$, $\left| \sum_{i=m}^n a_{f(i)} \right| = \sum_{i=m}^n |a_{f(i)}| < \epsilon. $ Therefore, any rearrangement will converges. And it's not hard to see they converges to the same sum.