Series Rearrangement Show that if the series A is absolutely convergent then rearrangement of the series is also convergent and converges to the same limit.
I can't think of any way to solve. Please help
 A: Suppose $A = \sum_{n=1}^{\infty} a_n$, where the convergence is absolute. Let $b : \mathbb{N} \to \mathbb{N}$ be a bijection.
Let $\epsilon > 0$. Since the series converges absolutely to $A$, there exists $N \in \mathbb{N}$ such that
$$\left|\sum_{n=1}^{M} a_n - A\right| < \epsilon/2$$
and
$$\sum_{n=M}^{\infty} |a_n| < \epsilon/2$$
for all $M \geq N$.
Let $M$ be large enough that $C_N = \{1,2,\ldots,N\} \subseteq \{b(1),b(2),\ldots,b(M)\} = B_M$. Then
$$\begin{aligned}
\left|\sum_{n=1}^{M} a_{b(n)} - A\right|
&= \left|\sum_{n \in B_M} a_{n} - A\right| \\
&= \left|\sum_{n \in C_N} a_n + \sum_{n \in B_M \setminus C_N}a_n - A \right|\\
&\leq \left| \sum_{n \in C_N} a_n - A\right| + \left| \sum_{n \in B_M \setminus C_N}a_n \right| \\
&= \left| \sum_{n=1}^{N} a_n - A \right| + \left| \sum_{n \in B_M \setminus C_N} a_n \right| \\
& \leq \epsilon/2 + \left| \sum_{n \in B_M \setminus C_N} a_n \right| \\
& \leq \epsilon/2 + \sum_{n \in B_M \setminus C_N} |a_n| \\
& \leq \epsilon/2 + \sum_{n=N+1}^{\infty}|a_n| \\
& \leq \epsilon
\end{aligned}$$
A: We use
Definition 1. $x$ and $y$ are $\varepsilon$-close iff $|x-y|\le\varepsilon$.
Lemma 2. Let  $\sum_{n=0}^\infty a_n$ be a convergent  series  of non-negative real  numbers,  and  let  $f\colon\mathbf N\to\mathbf N$  be  a  bijection.  Then $\sum_{m=0}^\infty a_{f(m)}$ is  also  convergent,  and  has  the  same  sum: $$\sum_{n=0}^\infty a_n=\sum_{m=0}^\infty a_{f(m)}.$$
(You can see the proof at the end of the answer.)
Lemma 3. Let $\sum_{n=m}^\infty a_n$ be a series of real numbers. Then $\sum_{n=m}^\infty a_n$ converges if and only if, for every real numer $\varepsilon>0$, there exists an integer $N\ge m$ such that $$\left|\sum_{n=p}^q a_n\right|\le\varepsilon\text{ for all }p,q\ge N.$$

Now we have prove
Proposition 4. Let $\sum_{n=0}^\infty a_n$ be an absolutely convergent series of real numbers, and let $f\colon\mathbf N\to\mathbf N$ be a bijection. Then $\sum_{m=0}^\infty a_{f(m)}$ is also absolutely convergent, and
has the same sum: $$\sum_{n=0}^\infty a_n=\sum_{m=0}^\infty a_{f(m)}.$$
Proof. We apply Lemma 2 to the infinite series $\sum_{n=0}^\infty |a_n|$, which by hypothesis is a convergent series of non-negative numbers. If we write $L:=\sum_{n=0}^\infty |a_n|$, then by Lemma 2 we know that $\sum_{m=0}^\infty |a_{f(m)}|$ also converges to $L$.
Now write $L':=\sum_{n=0}^\infty a_n$. We have to show that $\sum_{m=0}^\infty a_{f(m)}$ also converges to $L'$. In other words, given any $\varepsilon>0$, we have to find an $M$ such that $\sum_{n=0}^\infty |a_n|$ is $\varepsilon$-close to $L'$ for every $M'\ge M$.
Since $\sum_{n=0}^\infty |a_n|$ is convergent, we can use Lemma 3 and find an $N_1$ such that $\sum_{n=p}^{M'}a_{f(m)}\le\varepsilon/2$ for all $p,q\ge N_1$. Since $\sum_{n=0}^\infty a_n$ converges to $L'$, the partial sums $\sum_{n=0}^Na_n$ also converge to $L'$, and so there exists $N\ge N_1$ such that $\sum_{n=0}^Na_n$ is $\varepsilon/2$-close to $L'$.
Now the sequence $(f^{-1}(n))_{n=0}^N$ is finite, hence bounded, so there exists an $M$ such that $f^{-1}(n)\le M$ for all $0\le n\le N$. In particular, for any $M'\ge M$, the set $\{f(m):m\in\mathbf N; m\le M'\}$ contains $\{n\in N:n\le N\}$ (why?). So, for any $M'\ge M$
$$\sum_{m=0}^{M'}a_{f(m)}=\sum_{n\in\{f(m):m\in\mathbf N;m\le M'\}}a_n=\sum_{n=0}^Na_n+\sum_{n\in X}a_n$$
where $X$ is the set $$X=\{f(m):m\in\mathbf N;m\le M'\}\setminus\{n\in\mathbf N:n\le N\}.$$ The set $X$ is finite, and is therefore bounded by some natural number $q$; we must therefore have $$X\subseteq\{n\in\mathbf N: N+1\le n\le q\}$$ (why?). Thus $$\left|\sum_{n\in X}a_n\right|\le\sum_{n\in X}|a_n|\le\sum_{n=N+1}^q|a_n|\le\varepsilon/2$$ by our choice of $N$. Thus $\sum_{m=0}^{M'}a_{f(m)}$ is $\varepsilon/2$-close to $\sum_{n=0}^Na_n$, which as mentioned before es $\varepsilon/2$-close to $L'$. Thus $\sum_{m=0}^{M'}a_{f(m)}$ is $\varepsilon$-close to $L$ for all $M'\ge M$, as desired. $\quad\square$

Extra. Proof of Lemma 2.

 We  introduce  the  partial  sums  $S_N :=\sum_{n=0}^N a_n$  and $T_M:=\sum_{m=0}^Ma_{f(m)}$. We  know  that  the  sequences  $(S_N)_{n=0}^\infty$ and $(T_M)_{n=0}^\infty$ are  increasing. Write  $L:=\sup(S_N)_{n=0}^\infty$ and $L':=\sup(T_M)_{n=0}^\infty$. We  know  that  $L$ is finite,  and  in  fact  $L=\sum_{n=0}^Na_n$; we  see  that we will thus  be done as soon as we can show that  $L'=L$. Fix  $M$,  and  let  $Y$ be  the  set  $Y  :=\{m\in\mathbf N:  m\le M\}$.  Note that  $f$  is  a  bijection  between  $Y$  and  $f(Y)$.  We  have (why?) $$T_M=\sum_{m=0}^Na_{f(m)}=\sum_{m\in Y}a_{f(m)}=\sum_{n\in f(Y)}a_n.$$ The  sequence  $(f(m))_{m=0}^M$  is  finite,  hence  bounded,  i.e.,  there  exists an  $N$  such  that  $f(m)\le N$  for  all  $m\le N$.  In  particular  $f(Y)$  is a  subset  of  $\{ n  \in\mathbf  N  :  n\le N\}$,  and  so  (by the  assumption  that  all  the  an  are  non-negative) $$T_M=\sum_{n\in f(Y)}a_n\le \sum_{n\in\{n\in\mathbf N:n\le N\}}a_n=\sum_{n=0}^Na_n=S_N.$$ But  since  $(S_N)_{N=0}^\infty$  has  a  supremum  of $L$,  we  thus  see  that  $S_N\le L$, and  hence  that  $T_M\le L$  for  all $M$.  Since  $L'$  is  the  least  upper bound  of  $(T_M)_{M=0}^\infty$,  this  implies  that  $L'\le L$. A  very  similar  argument  (using  the  inverse  $f^{-1}$  instead  of  $f$) shows  that  every  $S_N$  is  bounded  above  by  $L'$,  and  hence  $L\le L'$. Combining  these  two  inequalities  we  obtain  $L  =  L'$,  as  desired.

