Show $\sum_{i=1}^\infty x_i y_i$ is absolutely convergent

Let $$\sum_{i=1}^\infty x_i$$ be absolutely convergent and let $$(y_n)$$ be a sequence satisying $$\forall i \in \mathbb N,\ \exists M \in \mathbb R$$ such that $$|y_i| \leq M$$. Then $$\sum_{i=1}^\infty x_i y_i$$ is absolutely convergent.

So $$\sum_{i=1}^\infty x_i$$ is absolutely convergent means both $$\sum_{i=1}^\infty x_i$$ and $$\sum_{i=1}^\infty |x_i|$$ is convergent. And the condition on y indicates y is bounded above.

For the proof, I have this idea: since if some series is convergent/divergent multiplication by a constant such as M does not change the convergence/divergence, I might be able to say that at worst case where $$y_i$$ is M, $$\sum_{i=1}^\infty x_i y_i$$ should have same convergent/divergent property with $$x_n$$. But I do not know how to show this proof formally.

A series $$\sum a_i$$ is absolutely convergent if $$\sum |a_i| <\infty$$. $$\sum |x_iy_i| \leq M\sum |x_i| <\infty$$. So $$\sum x_iy_i$$ is absolutely convergent. This is a complete proof.