Show that if ${ \sum_{n=1}^{\infty}{b_n}}$ absolutely convergent and ${ \sum_{n=1}^{\infty}{a_n} }$ convergent then the inner product ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ is also convergent.
Introductory Real Analysis by F. Dangello and M. Seyfried.
I feel like this could be proved using the Dirichlet's test but I did not know how to approach it from that direction. I also read something about Hadamard products on power series but I don't know if I could use it to prove my question. Furthermore, I read this question Convergence of product of convergent series but since I only have $b_n \ge 0$ the question is different.
My approach was to try to use the Cauchy condition stating that:
Since ${ \sum_{n=1}^{\infty}{a_n}}$ convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^na_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.
Furthermore, ${ \sum_{n=1}^{\infty}{b_n}}$ (absolutely)convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^nb_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.
Then, $\vert\sum_{n=1}^{\infty}{a_n}\vert \vert\sum_{n=1}^{\infty}{b_n} \vert = \vert \sum_{n=1}^{\infty}{a_n}{b_n} \vert \lt \sqrt\varepsilon\sqrt\varepsilon= \varepsilon$. Hence ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ convergent.
I feel like I made a mistake in the last line where I multiplied both series, and that my approach is too simplistic. So is my approach correct? Or where did I make a mistake? And how would I go about to prove this?
Best Answer
$\{a_n\}$ is eventually bounded by one as $\sum a_n$ is convergent. Hence for $N$ large enough and $n \ge N$
$$\vert a_n b_n \vert \le \vert b_n\vert$$
Which implies that $\sum a_n b_n$ is absolutely convergent as $\sum b_n$ is absolutely convergent.