Let ${a_n}_{n \in \mathbb Z_+}$ be a sequence of complex numbers. Suppose $Z_+ = \cup_1^{\infty} B_k$... 
Let ${a_n}_{n \in \mathbb Z_+}$ be a sequence of complex numbers. Suppose $\mathbb Z_+ = \cup_1^{\infty} B_k$ where each $B_k$ is infinite and $B_k \cap B_{k'} = \emptyset$ if $k \ne k'$. Prove that if $\sum_{k=1}^{\infty} \sum_{n \in B_k} \vert a_n \vert \lt \infty$, then $\sum a_n$ absolutely converges and is worth $\sum_{n =0}^{\infty} a_n = \sum_{k=1}^{\infty} \sum_{n \in B_k} a_n$.

Let's set $\phi_x : \mathbb Z_+ \to B_k$ bijection.
Given $n \in \mathbb N, \exists m \in \mathbb N$ such that
$\sum_{j=0}^n \vert a_j \vert \le \sum_{j=0}^{+\infty} \vert a_j \vert \lt + \infty \to \sum a_n$ absolutely converges.
Was all I could think of. Is it at least enough for the first part?
 A: I don't understand your logic for part 1; you are choosing $n$ and posing the existence of $m$ but don't use the latter variable. I think you are on the right track though, by bounding finite sums of $\sum |a_n|$. In particular, using the bijection $\phi : \mathbb{N}\to \mathbb{N}$ where $a_n \in B_{\phi(n)}$, we also have a well-defined function
$$\psi(n)=\max_{1\leq i\leq n}\phi(i),$$
which has the following properties:
$$\sum_{j=1}^n |a_j|\leq \sum_{j=1}^{\psi(n)}\sum_{k\in B_j}|a_k|;\quad (1)$$
$$\lim_{n\to\infty} \psi(n)=\infty,\quad (2)$$
where (1) holds because every term in the left is somewhere in the right. Taking limits as $n\to\infty$ and applying (2) we see that $\sum |a_n|$ is less than or equal than the sum that was given to converge, and so $\sum a_n$ absolutely converges.
For part 2, it intuitively feels like "rearrangement" question, since each term on the left has a corresponding term on the right. Of course any rearrangement presented as a single sum will equal the original sum because rearrangements of absolutely convergent series converge, and to the same value. But the right-hand sum is not strictly a "rearrangement". Maybe we can force it into one?
Define the doubly indexed sequence $a_{kn}'$ as
$$a_{kn}'=\begin{cases}a_n\quad & \text{if } a_n\in B_k \\ 0 &\text{otherwise}\end{cases}.$$
Then the double sum (the one on the RHS of part 2) is just
$$\sum_{k=1}^\infty \sum_{n=1}^\infty a'_{kn}.$$
If we can prove that the conditions of Fubini's theorem for infinite series holds for this case, then it can easily rearranged (note the order!) into
$$\sum_{n=1}^\infty \sum_{k=1}^\infty a'_{kn}=\sum_{n=1}^\infty a_n.$$
The easiest criterion (sufficient but not necessary) is that $\sum_{(n,k)\in \mathbb{N}\times \mathbb{N}}a'_{kn}$ is absolutely convergent, which, spelled out, means that for any $\epsilon>0$, there exists $K$ such that if $M,N\geq K$, then
$$\left|L-\sum_{n=1}^M\sum_{k=1}^N |a'_{kn}|\right|<\epsilon$$
for some limit $L$. Can you finish the proof from here?
