Prove that $\sum_{n,m=1}^{\infty}a_n b_m=(\sum_{n=1}^{\infty} a_n) (\sum_{m=1}^{\infty} b_m)$ Prove that if $\sum_{n=1}^{\infty} a_n$ and $\sum_{m=1}^{\infty} b_m$ are absolutely convergent series of real numbers then $$\sum_{n,m=1}^{\infty}a_n b_m=(\sum_{n=1}^{\infty} a_n) (\sum_{m=1}^{\infty} b_m)$$
is absolutely convergent.

Any hint please?
 A: On the left hand side 
$$\sum_{m,n=1}^\infty a_nb_m$$
I assume you are given some bijection $F: \mathbb N \to \mathbb N\times \mathbb N$ to define it. Then we write $\sum_k c_k$ to represent the series. 
Claim 1: $\sum_k c_k$ converges absolutely.
Proof: This is easy: for each $n\in \mathbb N$, there is $K$ so that 
$$F(\{1, 2, \cdots, n\}) \subset \{1, 2, \cdots , K\}^2.$$
So $$\sum_{k=1}^n |c_k| \le \sum_{m,n=1}^K |a_nb_m| \le \sum_{n=1}^\infty |a_n|\sum_{m=1}^\infty |b_m| <\infty. $$
Thus $\sum c_k$ converges absolutely. 
Claim 2: We have
$$\sum_{k=1}^\infty c_k = \sum_{n=1}^\infty a_n \sum_{m=1}^\infty b_m.$$
Proof: Let $\epsilon >0$. Then there is $N\in \mathbb N$ so that 
$$ \sum_{n=N+1}^\infty |a_n| , \sum_{m=N+1}^\infty |b_m| < \epsilon.$$
Let $N_\epsilon \in \mathbb N$ so that 
$$\{1, 2, \cdots, N\}^2 \subset F(\{1, 2, \cdots, N_\epsilon\}).$$
Then for all $n\ge N_\epsilon$, write $A_n \{ k: F(k) \notin \{1, 2, \cdots, N\}^2\} \cap \{1, 2, \cdots, n\}$.
$$\begin{split}
\left| \sum_{k=1}^n c_k - \sum_{n=1} ^\infty a_n \sum_{m=1}^\infty b_m\right|& =\left| \sum_{n=1} ^N a_n \sum_{m=1}^N b_m - \sum_{n=1} ^\infty a_n \sum_{m=1}^\infty b_m  + \sum_{k\in A_n} c_k\right| \\
&\le \left| \sum_{n=1} ^N a_n \sum_{m=1}^N b_m - \sum_{n=1} ^\infty a_n \sum_{m=1}^\infty b_m \right| + \sum_{k\in A_n} |c_k|\\
&= \left| \sum_{n=1} ^N a_n \sum_{m=N+1}^\infty b_m + \sum_{n=N+1} ^\infty a_n \sum_{m=1}^\infty b_m \right| + \sum_{k\in A_n} |c_k| \\
&\le \left( \sum_{n=1}^\infty |a_n| + \sum_{m=1}^\infty |b_m| \right)\epsilon + \sum_{k\in A_n} |c_k|
\end{split}$$
Similarly one has 
$$\sum_{k\in A_n} |c_k| \le  \left( \sum_{n=1}^\infty |a_n| + \sum_{m=1}^\infty |b_m| \right)\epsilon,$$
so 
$$\left| \sum_{k=1}^n c_k - \sum_{n=1} ^\infty a_n \sum_{m=1}^\infty b_m\right| \le 2\left( \sum_{n=1}^\infty |a_n| + \sum_{m=1}^\infty |b_m| \right)\epsilon$$
whenever $k\ge N_\epsilon$. So the claim is proved. 
