Product of absolutely convergent series and convergent series 
If $U=u_1+u_2+\ldots$ is an absolutely convergent series, and $V=v_1+v_2+\ldots$ is a convergent series, prove that $UV=u_1v_1+(u_1v_2+u_2v_1)+(u_1v_3+u_2v_2+u_3v_1)+\ldots$. 

So the series $|u_1|+|u_2|+\ldots$ converges, say, to $U_0$. That means for any $\varepsilon$, there exists $N$ such that for all $n>N$, $|u_n|+|u_{n+1}|+\ldots < \epsilon$. It follows that for any $\varepsilon$, there exists $N$ such that for all $m>n>N$, $|u_n|+|u_{n+1}|+\ldots+|u_m|<\varepsilon$ (this is also Cauchy's condition.)
Similarly, for any $\varepsilon$ there exists $P$ such that for all $m>n>P$, $v_n+v_{n+1}+\ldots+v_m<\varepsilon$. 
I tried to prove first that the series $u_1v_1+(u_1v_2+u_2v_1)+(u_1v_3+u_2v_2+u_3v_1)+\ldots$ converges by using the Cauchy criterion. But it seems difficult to bound $$(u_1v_n+u_2v_{n-1}+\ldots+u_nv_1)+(u_1v_{n+1}+u_2v_n+\ldots+u_{n+1}v_1)+\ldots$$
 A: This result is usually called "Mertens' theorem" (at least in French textbooks).
Here is one possible proof (the one that can be found in Rudin's "Principles of Mathematical Analysis").
Put $w_n=u_1v_n+\cdots +u_nv_1\, ,$
$U_n=u_1+\cdots+u_n$, $V_n=v_1+\cdots +v_n$ and $W_n=w_1+\cdots +w_n\, .$ It has to be shown that $W_n\to UV$ as $n\to\infty$. 
We have 
$$W_n=u_1V_n+\cdots +u_n V_1\, . $$
This can be written as 
$$W_n=\sum_{i=1}^\infty a_n(i)\, , $$
where 
$$a_n(i)=\left\{\begin{matrix}u_iV_{n+1-i}&{\rm if}\;i\leq n\\
0&{\rm if}\; i>n \end{matrix} \right. $$
For every fixed $i\in\mathbb N$, we have $\lim_{n\to\infty} a_n(i)=u_i V$. Moreover, $\vert a_n(i)\vert\leq C\, \vert u_i\vert$ for all $n,i$, where $C=\sup_k \vert V_k\vert$ (the sequence $(V_k)$ is bounded since it is convergent). Since the series $\sum u_i$ is absolutely convergent, one can apply the "dominated convergence theorem for series" to get that 
$$\lim_{n\to\infty}\sum_{i=1}^\infty a_n(i)=\sum_{i=1}^\infty \lim_{n\to\infty} a_n(i)\, , $$
in other words 
$$\lim_{n\to\infty} W_n= \sum_{i=1}^\infty u_iV=UV\, .$$
