Let $s_{nn}=\sum_{i=1}^n\sum_{j=1}^na_ib_j$, prove that the limit of it is $AB$

Question

THe problem first assumed that

$\sum_{i=1}^{\infty}\lvert a_i\rvert=A,\sum_{j=1}^{\infty}\lvert b_j\rvert=B$

It then asked me to prove:

$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\lvert a_ib_j\rvert$ converges

I did it. Now I am stuck in the problem.

$s_{nn}=\sum_{i=1}^n\sum_{j=1}^na_ib_j$, prove that the limit of it is $AB$

I approached it in several ways

1. Directly prove $s_{nn}$ converges to AB, but to no avail
2. Prove $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_ib_j$ converges to AB, but to no avail
3. Prove $\forall n\in\mathbb{N}\sum_{i=1}^n\sum_{j=1}^na_ib_j=(\sum_{i=1}^na_i)(\sum_{j=1}^nb_j)$, but then I am unable to prove the latter converges to $AB$.

I don't have a clue as to how to approach this now. The square summation is already a headache since it has $n^2$ term instead of the standard $n$ terms in series.

Edit 1: This is the problem from the book. I hope my translation to latex is accurate. In addition, this is copied from the same book yet different version. But it still gets the gist.

Edit 2: The definition of absolute convergence by the book.

Answer 1

After working on new assumption, I yield the following solution.

\begin{align} \forall n\in\mathbb{N}s_{nn}&=(\sum_{i=1}^na_i)(\sum_{j=1}^nb_j)\\ \rightarrow \lim_{n\to\infty}s_{nn}&=\lim_{n\to\infty}(\sum_{i=1}^na_i)(\sum_{j=1}^nb_j)\\ &=(\sum_{i=1}^{\infty}a_i)(\sum_{i=1}^{\infty}b_j)\\ &=AB \end{align}

Edit: Since no one objects, I will take it as my solution. :)