When multiply absolutely convergent seires and conditionally convergent series [closed]

Question

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?

Answer 1

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 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.