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.


1 Answer 1


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.


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