# Rearrangement Thm

Hi everyone: In the book what I've read of analysis in the proof of the Riemann rearrangement thm there is gaps that I need to fill. There is no real problem in almost everything but for the last one I have some troubles. I'll put all I have so far.

Rearrangement Thm: Let $\sum _{n=0}^\infty a_n$ be a series which is conditional convergent but not absolutely convergent, and let $L$ be any real number. Then there exists a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $\sum _{n=0}^\infty a_f(n)$ converges conditionally to $L$.

Proof: Let $A_-:=\{n\in \mathbb{N}: a_n< 0\}$ and $A_+:=\{n\in \mathbb{N}: a_n\ge 0\}$, it is no difficult to show that both sets are infinite. Furthermore, the series $\sum _{n\in A_-}a_n$, $\sum _{n\in A_+}a_n$ are absolutely divergent. Then we can find increasing bijections $f_-: \mathbb{N}\rightarrow A_-$ and $f_+: \mathbb{N}\rightarrow A_+$. And then, the series $\sum_{n=0}^\infty a_{f_-(n)}$, $\sum_{n=0}^\infty a_{f_+(n)}$ diverges absolutely.

Now we may define the sequence $n_0,n_1,..,n_j$. Suppose we have already defined $n_i$ for all $i<j$. We then define $n_j$ by the following rule:

Case I: If $\sum_{0\le i <j}a_{n_i}< L$ then we set $n_j:= \text{min} \{n\in A_+: \forall i<j.\ n\not=n_i\}$

Case II: If $\sum_{0\le i <j}a_{n_i}\ge L$ then we set $n_j:= \text{min} \{n\in A_-:\forall i<j.\ n\not=n_i\}$

The above definition is well-defined since the sets are nonempty.

Claim 1: The map $i \mapsto n_i$ is 1:1

Suppose that $k\not=j$. WLOG we may assume that $k<j$. Then either $\sum_{0\le i<j}a_{n_i}<L$ or $\sum_{0\le i<j}a_{n_i}\ge L$. For the former, $n_j:= \text{min} \{n\in A_+: \forall i<j.\ n\not=n_i\}$, i.e., $n_j \not= n_i$ for any $i<j$, in particular $n_j\not=n_k$. A similar argument shows that for the latter case that $n_j\not=n_k$. Hence, the map is 1:1.

Claim 2: The cases I and II occurs infinitely times.

For the sake of contradiction suppose that, say case II, occurs a finite number of times. Let $k$ be a sufficient large number such that for all $j>k$ the case II doesn't happen. Then $\sum_{i=0}^j a_{n_i}<L$ whatever $j>k$. Therefore $n_j\in A_+$ for all $j>k$. Moreover, $\sum_{i=0}^k a_{n_i}+\sum_{i=k+1}^j a_{n_i}<L$, so $\sum_{i=k+1}^j a_{n_i}<L-\sum_{i=0}^k a_{n_i}$ for any $j\ge k+1$ and also $a_{n_j}\ge 0$. Then the sequence of partial sums is bounded and hence is convergent.

Since $n_{k+1}\in A_+$, then there is some $\ell \in \mathbb{N}$ so that $f_+(\ell)=n_{k+1}$. We claim that for all $m\ge \ell$ there is some $j> k$, such that $f(m)=n_j$. The base case is when $m= \ell$ and is trivially true. Now suppose that we have proven the assertion for $m\ge \ell$. We know that $f_+(m+1)>f_+(m)=n_j$ (since is an increasing map) and also $\sum_{i=k+1}^j a_{n_i}<L-\sum_{i=0}^k a_{n_i}$, i.e., $\sum_{i=0}^j a_{n_i}<L$. Then $n_{j+1}= \text{min} \{n\in A_+:\forall i<j.\ n\not=n_i\}$. Now since $f_+(m+1)\not= n_j$ and clearly $f_+(m+1)\not= n_i$ for any $i< j+1$. Thus $f_+(m+1)\in \{n\in A_+: \forall i<j.\ n\not=n_i\}$ and is the minimum element. Thus $f_+(m+1)=n_{j+1}$ as desired.

Then for $M\ge k$ we have:

\begin{align}\sum_{i=0}^M a_{f_+(i)} &=\sum_{i=0}^k a_{f_+(i)}+\sum_{i=k+1}^M a_{f_+(i)}\\ &=\sum_{i=0}^k a_{f_+(i)}+\sum_{i=\ell}^{M*} a_{n_i}\\ &\le \sum_{i=0}^k a_{f_+(i)}+\bigg(L-\sum_{i=0}^k a_{n_i}\bigg) \end{align}

where $n_M*= f_+(M)$. Then, the series is bounded for any $M$ and thus is convergent, a contradiction. If we interchanged the roles of Case I and Case II a similar problem arises.

Claim 3: The map $i \mapsto n_i$ is a surjection.

By contradiction suppose there is some $i\in \mathbb{N} = A_+ \sqcup A_-$ such that $i\not= n_j$ for any $j$. Thus $i$ lies either in $\{n\in A_+: \forall i<j.\ n\not=n_i\}$ or $\{n\in A_-: \forall i<j.\ n\not=n_i\}$ but not in both at the same time for any $j$.

WLOG suppose that $i\in \{n\in A_+: \forall i<j.\ n\not=n_i\}$. Then, there is some $k\in \mathbb{N}$ such that $f_+(k)=i > f(k-1)$. If $i\not= n_j$ for any $j$ this means that Case $I$ don't occur $k+1$ times, and for what we have shown above this is not possible, then we reached a contradiction and the claim follows.

Claim 4: $a_{n_j}\rightarrow 0$

We already know that $a_n\rightarrow 0$. Let $\varepsilon>0$ be given. Let $N(\varepsilon)\ge 0$ such that $|a_n|\le \varepsilon$ for all $n\ge N(\varepsilon)$. Let define the set $\{i \in \mathbb{N}: n_i\le N(\varepsilon)\, \}$. Clearly the set is non-empty and also is bounded above, because is a finite set. Let $K$ be the supremum of the set. If $j> K$, then $n_j> N(\varepsilon)$. Then for all $j\ge K$ we have $|a_{n_j}| \le \varepsilon$ as desired.

Claim 5: $\text{Lim}_{j\rightarrow \infty} \sum_{0\le i <j}a_{n_i} =L$

For the sake of simplicity we denote $S_N= \sum_{i=0} ^N a_{n_i}$. Now suppose we have already defined $u_i, \ell_i$ for all $i<j$. We then define $u_j, \ell_j$ as follows:

$$u_j= \text{min}\{n\in \mathbb{N}: S_n\ge L, S_{n+1}< L, \text{ and } u_i\not= n \text{ for all } i<j\,\}$$

$$\ell_j= \text{min}\{n\in \mathbb{N}: S_n< L, S_{n+1}\ge L, \text{ and } \ell_i\not= n \text{ for all } i<j\,\}$$

The sequence is well-defined since both cases occurs infinitely many often, so the sets are never empty.

Then we have that $S_{\ell_i}<L$, $S_{\ell_i}+a_{n_{\ell_i+1}}\ge L$, ande $S_{u_i}\ge L$, $S_{u_i}+a_{n_{u_i+1}}< L$ for all $i$. So,

$$L-a_{n_{\ell_i+1}}\le S_{\ell_i}<L\le S_{u_i}< L-a_{n_{u_i+1}}$$

Since $a_{n_{\ell_i+1}},a_{n_{u_i+1}} \rightarrow 0$ (because is a subsequences of $a_{n_i}$), then by the squeeze thm we have $S_{\ell_i},S_{u_i}\rightarrow L$.

If $u_i < \ell_i$, then for all $u_i \le k \le \ell_i$ we therefore have $S_{\ell_i}\le S_k \le S_{u_i}$, thus converges to $L$. And if $\ell_i< u_i$, then for all $\ell_i\le k\le u_i$ we have $S_{\ell_i}\le S_k \le S_{u_i}$, thus converges to $L$.

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I have to say that your post affirms my belief that lots of notation often distracts from understanding. – Phira Dec 30 '13 at 20:41
I try to use the same notation that my book uses :P. – Jose Antonio Dec 30 '13 at 20:44
Yes, that is what I assumed. It is not forbidden to criticise books. :) – Phira Dec 30 '13 at 20:53
Indeed the book is amazing – Jose Antonio Dec 30 '13 at 21:53
I think, I need to modify some points, but the general idea I hope is correct. – Jose Antonio Dec 30 '13 at 22:17

Your partial sums cross the target value an infinite number of times. If you cross $L$ by adding a positive $a_{n_j}$ you cannot overshoot $L$ by more than $a_{n_j}$, if you cross it by adding a negative $a_{n_j}$, you also cannot get further from it than $a_{n_j}$. So after each crossing, you are stuck forever in a small interval around $L$ and since $a_{n_j}\to 0$, the sizes of the intervals tend to zero which means that the sum converges to $L$.