Absolute convergence of $\sum |x_n|$ and $\sum |y_n|$ implies absolute convergence of $\sum |x_ny_n|$

Question

Is the following argument correct?

Theorem. If $\sum x_n$ and $\sum y_n$ converge absolutely, then $\sum x_ny_n$ converges absolutely.

Proof. By hypothesis the series $\sum |x_n|$ and $\sum |y_n|$ are Cauchy, consequently given an arbitrary $ε>0$ we have $$\left|\sum_{j=k+1}^{n}|x_j|\right|<ε^2,\quad\forall k\ge M_1,\forall n>k$$ and $$\left|\sum_{j=r+1}^{l}|y_j|\right|<\frac1ε,\quad\forall r\ge M_2,\forall l>r$$ for some $M_1,M_2\in\mathbf{N}$. Now let $N = \max\{M_1,M_2\}$ and consider an arbitrary $p\ge N$ and $q>p$ we see that $$\left|\sum_{j=p+1}^{q}|x_j||y_j|\right| = \sum_{j=p+1}^{q}|x_j||y_j|\leq \left( \sum_{j=p+1}^{q}|x_j|\right)\left( \sum_{j=p+1}^{q}|y_j|\right)<ε^2\cdot\frac{1}{ε} = ε,$$ thus the series $\sum |x_ny_n|$ is Cauchy and thus converges absolutely.

$\blacksquare$

Answer 1

Your proof is correct. Let me also suggest an alternate proof. Convergence of $\sum y_n$ implies $y_n \to 0$ so $|y_n|<1$ for $n$ sufficiently large. Hence $|x_n||y_n| \leq |x_n|$ for $n$ sufficiently large from which the result follows.

Answer 2

Your proof is correct.

Hölder's inequality with $p=1,q=\infty$ implies the statement:

$$\sum_{i=1}^{\infty}|x_ny_n|={\Vert xy\Vert}_{l^1}\leq {\Vert x\Vert}_{l^1}{\Vert y\Vert}_{l^{\infty}}=\left (\sum_{i=1}^{\infty}|x_n|\right )\sup_{n\in\mathbb{N}}|y_n|<\infty$$ $\sup_{n\in\mathbb{N}}|y_n|<\infty$ holds. Otherwise $\left ( \sum_{i=1}^{k}|y_n|\right )_{k\in\mathbb{N}}$ would not converge.