Complex analysis (theorem related to properties of Cauchy product of complex series) - request for proof-explanation Theorem: Let: $\sum\limits_{n=1}^{+\infty}a_{n}$,  $\sum\limits_{n=1}^{+\infty}b_{n}$ - conditionally convergent complex series,   $\sum\limits_{n=1}^{+\infty}c_{n}$ - Cauchy product of $\sum a_n$, $\sum b_n$ such that $\sum c_n$ converges. Then: $\sum\limits_{n=1}^{+\infty}c_{n} = (\sum\limits_{n=1}^{+\infty}a_{n})(\sum\limits_{n=1}^{+\infty}b_{n})$
Proof: Let: $A_n$,$B_n$,$C_n$ - partial sums of $\sum a_n$, $\sum b_n$, $\sum c_n$ respectively, $A_n \to A$, $B_n \to B$,  $C_n \to C$.
Define: $\alpha_{n}:=A_n-A$, $\beta_n:=B_n-B$, $p_n:=\frac{\sum\limits_{i=0}^{n}\alpha_i}{n}$,  $q_n:=\frac{\sum\limits_{i=0}^{n}\beta_i}{n}$
Algebraic manipulations imply: $\frac{\sum\limits_{i=0}^{n-1}C_i}{n}=AB+Aq_n+Bp_n+r_n$ (*)
, where: $r_n:=\frac{\sum\limits_{i=0}^{n-1}\alpha_i\beta_{n-i}}{n}$
But: $\lim\limits_{n\to +\infty}(\alpha_n) = 0 \space \wedge \lim\limits_{n\to +\infty}(\beta_n) = 0 \implies \lim\limits_{n\to +\infty}(p_n) = 0 \space \wedge \lim\limits_{n\to +\infty}(q_n) = 0 \space \wedge \lim\limits_{n\to +\infty}(r_n) = 0$ 
Hence: right hand side of (*) tends to $AB$. 
Also: $\lim\limits_{n\to +\infty}(C_n) = C$. Therefore: $C=AB$ $\square$
What I do not understand in this proof: 


*

*I completely do not know how to derive equation labelled (*). Method of direct substitution of alternative form of symbols completely fail. There must be some trick in order to obtain this equation, however I can't find it. 

*I do not see why $\lim\limits_{n\to +\infty}(r_n) = 0$.
I would be very thankful to anyone that would post explanation towards these 2 steps in the above proof. Apart from these issues, other parts of this proof are clear for me.
 A: So the goal is to show that, if $\sum_{n=0}^\infty a_n = A$ and $\sum_{n=0}^\infty b_n = B$ are two conditionally convergent complex series, and $c_n = \sum_{i=0}^n a_i b_{n-i}$ is the Cauchy product of $a_n$ and $b_n$ and the series $\sum_{n=0}^\infty c_n$ converges to a value $C$, then $C = AB$.
First let us notate the the partial sums of $a_n$, $b_n$ and $c_n$ as $A_n$, $B_n$ and $C_n$ respectively. Then to prove the theorem we will attempt to use the fact that, since by assumption $C_n$ converges to $C$
$$
 C = \lim_{n\to\infty} C_n = \lim_{N\to\infty} \frac1N \sum_{n=0}^N C_n.
$$
The first step of the proof is then to show that
$$
\begin{align}
\frac1N \sum_{n=0}^N C_n 
&= \frac1N \sum_{n=0}^N \sum_{k=0}^n \sum_{i=0}^k a_i b_{k-i} \\
&= \frac1N \sum_{n=0}^N \sum_{\substack{i + j \leq n}} a_i b_{j} \\
&= \frac1N \sum_{n=0}^N \sum_{i=0}^n a_i \sum_{j=0}^{n-i} b_{j} \\
&= \frac1N \sum_{n=0}^N \sum_{i=0}^n a_i B_{n-i} \\
&= \frac1N \sum_{\substack{i + j \leq N}} a_i B_j \\
&= \frac1N \sum_{i=0}^N A_i B_{N-i}
\end{align}
$$
from which it is relatively easy to show that
$$
\begin{align}
\frac1N \sum_{n=0}^N C_n 
&= \frac1N \sum_{i=0}^N A_i B_{N-i}\\
&= \frac1N \sum_{i=0}^N (A + \alpha_i) (B + \beta_{N-i})\\
&= \frac1N \sum_{i=0}^N AB + \alpha_i B + A \beta_{N-i} + \alpha_i \beta_{N-i}\\
&= AB + \left(\frac1N \sum_{i=0}^N \alpha_i\right)B + A \left(\frac1N \sum_{i=0}^N \beta_{N-i}\right) + \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i}\\
\end{align}
$$
(this should answer your first question).
Now because $a_n \to 0$ and $b_n \to 0$ (as $n \to \infty$) it follows that $\frac1N \sum_{i=0}^N \alpha_i \to 0$ and $\frac1N \sum_{i=0}^N \beta_i \to 0$ as well. Therefore
$$
\begin{align}
C 
&= \lim_{N\to\infty} AB + \left(\frac1N \sum_{i=0}^N \alpha_i\right)B + A \left(\frac1N \sum_{i=0}^N \beta_{N-i}\right) + \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i} \\
&=AB + \left( \lim_{N\to\infty}  \frac1N \sum_{i=0}^N \alpha_i\right)B + A \left( \lim_{N\to\infty}  \frac1N \sum_{i=0}^N \beta_{i}\right) +  \lim_{N\to\infty}  \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i} \\
&= AB + \lim_{N\to\infty} \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i}
\end{align}
$$
so it remains to show that $\lim_{N\to\infty} \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i} = 0$. For this I haven't been able to find a particularly elegant proof, but note that $|\alpha_n| \to 0$ (and also $|\beta_n| \to 0$) as $n \to \infty$ so in particular the sequence $|\alpha_n|$ can't diverge, hence it has an upper bound $U$. And therefore
$$
\left| \frac1N \sum_{i=0}^N \alpha_i \beta_{N-i} \right| \leq  \frac1N \sum_{i=0}^N |\alpha_i \beta_{N-i}| \leq  \frac1N \sum_{i=0}^N U |\beta_{N-i}|  = U \frac1N \sum_{i=0}^N |\beta_{i}| 
$$
And $|\beta_n| \to 0$ implies that
$$
\lim_{N\to\infty} \frac1N \sum_{i=0}^N |\beta_{i}| = 0
$$
completing the proof.

Addendum: I suspect that there might be a proof that doesn't use the convergence of $\alpha_n$ and $\beta_n$ as there is a very similar theorem when $a_n$ and $b_n$ are merely Cesàro summable (meaning their series may not even converge conditionally), which we've almost proven, but I think this will require some non-trivial effort. Although it is also possible that you need the theorem we just proved to prove the more general version.
