Show that if $(\sum x_n)$ converges absolutely and $(y_n)$ is bounded then $(\sum x_n y_n)$ converges

Question

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 $$\sum_{n=m}^{t} |x_n|<\frac{\varepsilon}{B}\implies\left|\sum_{n=m}^{t} x_n y_n\right|\le B\sum_{n=m}^{t}|x_n|<\varepsilon$$

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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 $(\sum \frac1n)$

Answer 1

(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.