Is this series absolutely converges
$$
\sum\_{n=1}^{\infty}C\_nU\_n
$$
when


$\sum\_{n=1}^{\infty}C\_n$ is absolutely convergent series and


$\sum\_{n=1}^{\infty}U\_n$ is conditionally convergent series?



When multiply absolutely convergent seires and conditionally convergent series
I don't think this could be a considered a research level question: perhaps, this could be defined a ~~tricky~~ simple exercise. However, the answer is yes as I am going to show below


If $\sum\_{n=1}^\infty C\_n$ is absolutely convergent, then we have that its [total variation as a sequence](https://en.wikipedia.org/wiki/Bounded_variation#bv_sequences) is bounded, i. e.
$$
\sum\_{n=1}^\infty |C\_n| <\infty \implies TV(C\_n)=\sum\_{n=1}^\infty |C\_{n+1} - C\_n|<\infty
$$
since
$$
\infty >\sum\_{n=1}^\infty |C\_n|+ \sum\_{n=1}^\infty |C\_{n+1}|= \sum\_{n=1}^\infty \big[|C\_n|+ |C\_{n+1}|\big]\ge \sum\_{n=1}^\infty |C\_{n+1} - C\_n| \ge 0
$$
Now you can apply the *du Bois-Reymond - Dirichlet test* ([1], chapter X, §43 p. 315) and conclude the result you asked for.


**Addendum**


After the comment Dieter Kadelka, I noticed that his solution is even more straightforward and stronger. To explain it, note that the convergence of $\sum\_{n=1} U\_n$ (conditionally or not) implies $\lim\_{n\to\infty} U\_n= 0$ (i.e. $(U\_n)\_{n\in\Bbb N\_+} \in c\_0$) thus $(U\_n)\_{n\in\Bbb N\_+}$ is bounded $\iff |U\_n|\le M$ for all ${n\in\Bbb N\_+}$ and a given $M>0$. his finally implies
$$
\left|\sum\_{n=1}^\infty U\_n C\_n\right|\le M \sum\_{n=1}^\infty|C\_n|
$$
and thus the product series converges absolutely.


**Reference**


[1] Konrad Knopp, *Theory and application of infinite series*, translated from the 2nd edition and revised in accordance with the fourth by R. C. H. Young. (English)
London-Glasgow: Blackie & Son, Ltd. XII, 563 p. (1951), [Zbl 0042.29203](https://zbmath.org/?q=an%3A0042.29203).


