Proof about Theorem of Dirichlet. Good morning,
Help with the proof of this theorem
If $\sum_{n=1}^{\infty}a_{n}$ is Absolutely convergent, then any rearrangement of 
$\sum_{n=1}^{\infty}a_{n}$ absolutely convergent and converge to the same value of $\sum_{n=1}^{\infty}a_{n}$
I have some ideas, but nothing clear. For example: 
Because $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent, Then $\sum_{n=1}^{\infty}a_{n}=S$ In other words,
$\lim_{n\rightarrow\infty}S_{n}=\lim_{n\rightarrow\infty}(|a_{1}|+|a_{2}|+...+|a_{n}|)=S\in\mathbb{R}$ 
Let
$\sum_{n=1}^{\infty}b_{n}$ a rearrangement of $\sum_{n=1}^{\infty}a_{n}$, in this step i'm stuck because if $S_{n}$ converge then $T_{n}$(Partial Sum of serie $b_{n}$) should converge, but i cannot prove that.
Can someone help?
 A: Suppose at first that all the terms of series are non-negative. Let $\sigma: \mathbb{N} \mapsto \mathbb{N}$ be a permutation, so that 
$$ \sum_{n \ge 1} b_n = \sum_{n \ge 1} a_{\sigma(n)} $$
is a rearrangement of the initial series. Let $\alpha_n , \beta_n $ the sequences of the partials sums of $\sum a_ n, \sum b_n $ respectively. For every $n \in \mathbb{N}$, let:
$$ k_n = \max\{\sigma(p) : 0 \le p \le n \}  $$
Since every series involved has non-negative terms it holds:
$$ \beta_n = \sum_{i=1}^{n} b_i =\sum_{i=1}^{n} a_{\sigma(i)} \le \sum_{j=0}^{k_n } a_j = \alpha_{k_n} \le a  $$
which means that the partials sums $\beta_n $ are monotone and bounded above, hence convergent to a limit $b$ which is less or equal then $a$. Noting that we can see the initial series as a rearrangement of the rearranged one, we also get $b \ge a$, whence $a=b$.
If the series $\sum_{n \ge 1} a_n$ has terms with non-constant sign, we still get the same conclusion thanks to the positive and negative part decomposition, and noting that $\sum (b_n)^{+}$ and $\sum_{n \ge 1} (b_n)^{-}$ are rearrangements of $\sum_{n \ge 1} (a_n)^{+}$ and $\sum_{n \ge 1} (a_n)^{-}$, respectively.  
A: Hint: Define $$c_i=\begin{cases}0&a_i<0\\a_i&a_i\geq 0\end{cases}\\
d_i = \begin{cases}0&a_i>0\\-a_i&a_i\leq 0\end{cases}$$
Show that $c_i$ and $d_i$ are non-negative, that $c_i+d_i=|a_i|$ and $c_i-d_i=a_i$, and, if $\sum a_i$ is absolutely convergent, then $\sum c_i$ and $\sum d_i$ are convergent.
Now use this to show that any re-arrangement of $\sum a_i$ must converge to the same value.
A: First, let's deal with the case where $a_n\geq 0$ for all $n\in\mathbb{N}$.
Let $(b_n)_{n\in\mathbb{N}}$ be a rearrangement of $(a_n)_{n\in\mathbb{N}}$, and let $\epsilon>0$ be given.  Because $\sum_{n=1}^{\infty}a_n$ converges, its tail sums must converge to $0$; so, there exists $N\in\mathbb{N}$ so that $k>N$ implies
$$
0\leq \sum_{n=k}^{\infty}a_n<\epsilon.
$$
Now, choose $M$ large enough that the elements $a_1,\ldots,a_N$ are all accounted for in the rearrangement by $b_1,\ldots,b_M$.
For any $m>M$, let's consider the differences between the partial sums $\sum_{n=1}^{m}a_n$ and $\sum_{n=1}^{m}b_n$.  The terms $a_1,\ldots, a_N$ occur in both.  If we cancel those, what's left? Use the triangle inequality to write
$$
\left\lvert\sum_{n=1}^{m}a_n-\sum_{n=1}^{m}b_n\right\rvert\leq A+B,
$$
where $A$ is the sum of left-over terms from the first sum and $B$ from the second.  Then
$$
0\leq A\leq\sum_{n=N+1}^{m}a_n\leq\sum_{n=N+1}^{\infty}a_n<\epsilon,
$$
since we've canceled out the terms $a_1,\ldots,a_N$.  What about $B$?  The right sum consisted of $a_1,\ldots,a_N$ (out of order), plus extra terms; so, any extra terms must correspond to elements $a_k$ from the original series such that $k>N$. So, the same inequalities yield $0\leq B<\epsilon$.
So, given any definition of $\epsilon$, by taking $m$ large enough we can force the partial sums of the first $m$ terms of each series to differ by at most $2\epsilon$; hence, the difference between the sequences of partial sums converges to $0$ as $m\to\infty$!  So, since the original series converged, the rearrangement must also converge to the same value.
Now, what if not all terms are positive?  Show that the sum of positive terms and the sum of negative terms must each converge, and that therefore you can break the series up into these two sums.  Then, use this to argue that the result still holds.
A: Let 
$$S_n=a_1+...+a_n \\
T_n=b_1+..+b_n=a_{\sigma(1)}+...+a_{\sigma(n)}$$
where $b_n$ is a rearrangement of $a_n$.
Let $\epsilon >0$. Since the original series is absolutely convergent, there exists some $N$ so that 
$$\sum_{n >N} |a_n| <\epsilon$$
Next, pick $M$ such that the set $\{ \sigma(1),.., \sigma(M) \}$ contains the set $\{1, 2,..., N \}$. Such an $M$ exists because $\sigma$ is a bijection.
Next, if $n > \max \{ M,N \}$ then all the terms $a_1,..,a_N$ appear in both $S_n$ and $T_n$, and therefore, they cancel in $S_n-T_n$.
This shows that for all $n > \max \{ M,N \}$ we have
$$|S_n -T_n | \leq \sum_{n >N} |a_n| <\epsilon$$
Therefore, we showed with $\epsilon, N$ that 
$$\lim_{n \to \infty} S_n-T_n=0$$
The claim follows.
