I've spent a lot of time working this problem in the following way. Have I significantly messed my logic up at some point? Also, I've went to office hours to discuss this partial solution to my professor, but he was busy so he didn't get to help me as much as I would have like. If my argument is sound up to this point, then could someone please help me move forward in the direction I am currently going in?

Problem: If the sequence $\{y_k\}$ is bounded and $\sum |x_k|$ converges, then $\sum x_k y_k$ converges.

Proof: Let $S_n = x_1 y_1 + x_2 y_2 + \cdots + x_n y_n$ be the sequence of partial sums. We must now show that $S_n$ converges which will imply that $\sum x_k y_k$ converges. However, let us first recall the following: \begin{align*} \sum |x_k| \text{ is convergent} \Rightarrow |x_k|\rightarrow 0\Rightarrow x_k\rightarrow 0\\ \end{align*} $$\text{Moreover, if } \{y_k\} \text{ is bounded and } x_k\rightarrow 0, \text{ then }x_k y_k\rightarrow 0.\\$$ Now, we say $S_n$ converges if and only if $\forall \varepsilon >0 \, \exists n_0 \in \mathbb{N}$ such that if $n>m\ge n_0$, then $|x_{m+1}y_{m+1}+x_{m+2}y_{m+2}+\cdots+x_{n}y_{n}|<\varepsilon$. Yet, by the triangle inequality, $$|x_{m+1}y_{m+1}+x_{m+2}y_{m+2}+\cdots+x_{n}y_{n}|\le |x_{m+1}y_{m+1}|+|x_{m+2}y_{m+2}|+\cdots+|x_{n}y_{n}|.$$

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    $\begingroup$ So far so good. Showing that $\sum x_n y_n$ converges absolutely is the way to go. Now use that $(y_n)$ is bounded, and $\sum x_n$ absolutely convergent. $\endgroup$ – Daniel Fischer Oct 6 '14 at 18:39
  • $\begingroup$ Thank you for the help and the confident booster. $\endgroup$ – Valentino Oct 6 '14 at 19:15

The sequence $\{y_n\}$ is bounded so there exists a finite number $M \geq 0$ such that $|y_n| \leq M$ for all $n$.

Therefore, $|\sum_n x_n y_n| \leq \sum_n |x_n y_n| = \sum_n |x_n||y_n| \leq M \sum_n |x_n| < \infty$, as needed.

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  • $\begingroup$ Thank you for this solution. $\endgroup$ – Valentino Oct 6 '14 at 19:14

Since $\{y_n\}$ is bounded there is a positive number $M$ such that $|y_n|<M$ for every $n \in \mathbb{N}$

Let $\varepsilon>0$ be given, and set $s_n=|x_1|+|x_2|+\cdots+|x_n|$.

Since the series $\sum \limits_{n=1}^\infty |x_n|$ converges we may let $s=\lim_{n \to \infty}s_n$. By definition there is $N \in \mathbb{N}$ so that $|s_n-s|<\varepsilon/M$ whenever $n \geq N$.

Notice that $M\sum \limits_{n=1}^\infty |x_n|=\sum \limits_{n=1}^\infty M|x_n|$.

Now $|Ms_n-Ms|=M|s_n-s|<\varepsilon$ whenever $n \geq N$, so we have that the series $\sum \limits_{n=1}^\infty M|x_n|$ converges.

Since $|y_n x_n| \leq M|x_n|$ for all positive integers $n$, the series $\sum \limits_{n=1}^\infty |y_n x_n|$ converges by the comparison test.

Thus $\sum \limits_{n=1}^\infty y_n x_n$ converges, as it is an absolutely convergent series.

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