Absolute convergence, interpretation of summation? Suppose $\sum_{n=0}^\infty a_n$ and $\sum_{m=0}^\infty b_m$ converge absolutely. I have to show that $$\left(\sum_{n=0}^\infty a_n\right) \cdot \left(\sum_{m = 0}^\infty b_m\right) = \sum_{m, n}^\infty a_nb_m.$$ But I do not understand what the sum on the right-hand side means (i.e. what limit this represents). Could anyone help explain it?
 A: You are perfectly right in non-understanding what this "double sum" means.
Here is one possible interpretation of what you have to prove. This relies on the notion of summability for a family of real numbers. Let $(c_i)_{i\in I}$ be a family of real numbers indexed by some set $I$. Then, $(c_i)_{i\in I}$ is said to be summable if there exists a real number $S$ such that the following holds : for any $\varepsilon>0$, one can find a finite set $F\subset I$ such that $\vert \sum_{i\in F'} c_i-S\vert<\varepsilon$ for any finite set $F'\subset I$ containing $F$. In this case, $S$ is uniquely determined, and denoted by $S:=\sum_{i\in I} c_i$. In your situation, the index set is $\mathbb N\times \mathbb N$ and $c_i=a_nb_m$ for $i=(n,m)\in\mathbb N\times \mathbb N$. So, what you have to prove could read as follows: show that the family $(a_nb_m)_{(n,m)\in\mathbb N\times \mathbb N}$ is summable with sum the product of the two sums on the left-hand side.
As it turns out, if the index set $I$ is countable, then a family $(c_i)_{i\in I}$ is summable with sum $S$ if and only if, for every bijection $\phi:\mathbb N\to I$, the series $\sum c_{\phi(n)}$ is convergent with $\sum_1^\infty c_{\phi (n)}=S$. In other words, if you enumerate (in a 1-1 way) the set $I$ as $\{ i_n;\; n\in\mathbb N\}$ then the series $\sum_n c_{i_n}$ should be convergent with $\sum_0^\infty c_{i_n}=S$, independently of the enumeration you have chosen. This is also equivalent to absolute convergence of $\sum c_{i_n}$ for some enumeration of $I$ (and then this holds for every enumeration). So, what you are asked to do may be the following: show that for any enumeration of the set $\mathbb N\times \mathbb N$ by the integers, i.e. whenever you write $\mathbb N\times \mathbb N$ in a 1-1 fashion as $(n_k,m_k)_{k\in\mathbb N}$, the series $\sum a_{n_k}b_{m_k}$ is convergent with sum the product of the two sums in the left-hand side.
Another possibility is that you are "just" asked to prove the standard theorem on "product series"; namely, if you set $c_k=\sum_{n=0}^k a_nb_{k-n}$ then the series $\sum c_k$ is convergent with sum the product of the two sums in the left-hand side. This corresponds to the "usual" enumeration of $\mathbb N\times \mathbb N$ obtained by going through each "diagonal" $\Delta_k=\{ n+m=k\}$, $k\in\mathbb N$.
A: The notation $$\sum_{m, n = 0}^\infty a_nb_m$$ should be interpreted as the sum $$\sum_{k=1}^\infty a_{n_k}b_{m_k}$$ where for each ordered pair $(m, n)$ of nonnegative integers, there exists $k$ such that $(m_k, n_k) = (m, n)$. The sum is not well-defined in general. Since we are not given a way to order the ordered pairs $(m_k, n_k)$, the sum can only be well defined if it does not depend on the order of summation. One is supposed to prove this by showing the sum converges absolutely. In general, if one sees a sum of terms indexed by an infinite set, one must interpret it this way unless one is given an order of summation.
It is an instructive exercise to show one can rearrange the sum on the right-hand side of the original equation as follows:$$\sum_{\ell = 0}^\infty \left(\sum_{m + n = \ell} a_nb_m\right).$$
