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

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.

$$\blacksquare$$

• seems fine. I simpler proof is noting that if $\sum_n |x_n|$ converges then there is some $N\in\Bbb N$ such that $|x_n|<1$ for all $n\ge N$. Then clearly $|x_n y_n|\le|y_n|$ for all $n\ge N$, hence by comparison test the series $\sum_n |x_n y_n|$ also converges – Masacroso Nov 26 '18 at 8:59
• @Masacroso would not be $|x_n|\le1$ ? – JD_PM Feb 7 at 15:12
• @JD_PM it could be $|x_n|\le 1$ also – Masacroso Feb 7 at 15:19

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.
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.