How this absolutely convergent series property work? An absolutely cconvergent series may be multiplied with another absoultely convergent series. The limit of the product will be the product of the individual series limits.
How does it work?
 A: Consider absolutely convergent series $\sum a_i$ and $\sum b_i$, for $i \ge 0$.  The product series is defined to be
$$
\sum_{n \ge 0} \left( \sum_{i=0}^n a_i b_{n-i}  \right)
$$
Proof that this series is absolutely convergent
For any $N$, we have
\begin{align*}
\sum_{n = 0}^N \left| \sum_{i=0}^n a_i b_{n-i} \right|
&\le \sum_{n = 0}^N \sum_{i=0}^n \left| a_i b_{n-i} \right| \\
&\le \sum_{i=0}^N \sum_{j=0}^N |a_i| |b_j| \\
&= \left( \sum_{i=0}^N |a_i| \right) \left( \sum_{j=0}^N |a_j| \right) \\
\end{align*}
which is uniformly bounded since $\sum |a_i|$ and $\sum |b_i|$ are uniformly bounded.
Proof that the series converges to the product
We compare the $2N$th partial sum with the product $\left(\sum_{i=0}^N a_i\right) \left( \sum_{i=0}^N b_i \right)$.
\begin{align*}
\left| \sum_{n=0}^{2N} \sum_{i=0}^n a_i b_{n-i}
- \left(\sum_{i=0}^N a_i\right) \left( \sum_{i=0}^N b_i \right) \right|
&= \left| \sum_{i, j}_{i + j \le 2N} a_i b_j - \sum_{i, j}_{i, j \le N} a_i b_j \right| \\
&= \left| \sum_{i, j}_{i + j \le 2N}_{i,j > N} a_i b_j \right| \\
&\le \sum_{i, j}_{i + j \le 2N}_{i,j > N} \left| a_i b_j \right| \\
&\le \sum_{i=0}^N \sum_{j=N+1}^{2N} |a_i b_j| + \sum_{i=N+1}^{2N} \sum_{j=0}^{N} |a_i b_g| \\
&\le
\left( \sum_{j=N+1}^{2N} |b_j| \right) \sum_{i=0}^\infty |a_i|
+ \left( \sum_{i=N+1}^{2N} |a_i| \right) \sum_{j=0}^\infty |b_j| \\
&\le
\left( \sum_{j=N+1}^{\infty} |b_j| \right) \sum_{i=0}^\infty |a_i|
+ \left( \sum_{i=N+1}^{\infty} |a_i| \right) \sum_{j=0}^\infty |b_j|
\end{align*}
Now taking the limit as $N \to \infty$, the terms in parentheses go to $0$
because they are the tail ends of convergent series.  Thus
$$
\lim_{N \to \infty} \sum_{n=0}^{2N} \sum_{i=0}^n a_i b_{n-i}
= \lim_{N \to \infty} \left(\sum_{i=0}^N a_i\right) \left( \sum_{i=0}^N b_i \right)
$$
(Both limits exist, since the latter limit is the product of limits which exist.)  Thus the product series converges absolutely to the product of the two series.
