I have come across the following problem and I believe I have a correct solution, but I had a question about the necessity of the hypotheses:
Prove that if $\sum_{n = 1}^{\infty} a_n$ and $\sum_{n = 1}^{\infty} b_n$ are absolutely convergent series, then $\sum_{n, m = 1}^{\infty} a_n b_m$ is absolutely convergent and converges to \[ \left(\sum_{n = 1}^{\infty} a_n \right) \left( \sum_{n = 1}^{\infty} b_n \right). \]
I was able to prove this by simply writing the sum of the products as a double nested sum and factoring out the inner sum, however the only part of my proof that depends on the series being absolutely convergent is to show that $\sum_{n, m = 1}^{\infty} a_n b_m$ is absolutely convergent. Is correct?
Namely, is it true that for any convergent series the sum of their products converges to the product of their sums, or did I make a mistake somewhere in my proof?
Without absolute convergence, what you mean by the notation
$$\sum_{n,m=1}^\infty a_n b_m$$
is ill-defined. For absolute convergence, we can say: define some bijection between
$$\{(n,m):n,m\in\mathbb{N}\}\leftrightarrow \mathbb{N}$$
by $s_k$ with the notation
$$s_k(1)=x\text{ in }(x,y)$$
$$s_k(2)=y\text{ in }(x,y)$$
Then
$$\sum_{n,m=1}^\infty a_n b_m=\sum_{k=1}^\infty a_{s_k(1)} b_{s_k(2)}=L$$
for a specific $L$ regardless of which bijection is used above. If one of the sums is conditionally convergent (or both) then you can rearrange the terms to get any limit you want, including $-\infty$ and $\infty$. Thus, for a conditionally convergent series there exists a bijection
$$\{(n,m):n,m\in\mathbb{N}\}\leftrightarrow \mathbb{N}$$
(called $s_k$) such that
$$\sum_{n,m=1}^\infty a_n b_m=\sum_{k=1}^\infty a_{s_k(1)} b_{s_k(2)}=L$$
where $L$ is any real number of positive/negative infinity.