This is the exercise 2.7.6 of the book Understanding analysis of Abbott, I want a check of my proof and if is needed additional information to complete it.
a) Show that if the sequence $(\sum x_n)$ converges absolutely and the sequence $(y_n)$ is bounded then the sequence $(\sum x_n y_n)$ converges
b) Find a counterexample that demonstrates that a) does not always hold if the convergence of $(\sum x_n)$ is conditional
For the part a): if $(\sum x_n)$ converges absolutely it means, using the definition of Cauchy sequences (that demonstrates convergence on complete spaces) applied to series that
$$\forall \varepsilon> 0, \exists N\in\Bbb N :\sum_{n=m}^{t} |x_n|<\varepsilon, \forall t>m> N$$
and we have the inequalities $|\sum x_n y_n|\le \sum|x_n y_n|$ and $|\sum x_n|\le\sum|x_n|$. And cause $(y_n)$ is bounded we have that $|y_n|\le B$ and then
$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|\le B\sum|x_n|$$
Then we have that $$\sum_{n=m}^{t} |x_n|<\frac{\varepsilon}{B}\implies\left|\sum_{n=m}^{t} x_n y_n\right|\le B\sum_{n=m}^{t}|x_n|<\varepsilon$$
For the part b) we can take the conditional convergent sequence $(\sum\frac{(-1)^{n-1}}{n})$ and the bounded sequence $((-1)^{n-1})$. If we multiply both as the problem define we get the divergent sequence $(\sum \frac1n)$
Best Answer
(Writing this just so the question can get marked as answered.) You are correct in both cases. You're right in the first part that it basically boils down to
$$\left|\sum\limits_{n}x_n y_n\right|\leq \sum\limits_{n}|x_n y_n| = \sum\limits_n |x_n||y_n|\leq \sup\limits_{k}|y_k|\sum\limits_{n}|x_n|< +\infty.$$
Your counter-example in the second part is good.