3
$\begingroup$

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.

$\blacksquare$

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Masacroso Nov 26 '18 at 8:59
  • $\begingroup$ @Masacroso would not be $|x_n|\le1$ ? $\endgroup$ – JD_PM Feb 7 at 15:12
  • 1
    $\begingroup$ @JD_PM it could be $|x_n|\le 1$ also $\endgroup$ – Masacroso Feb 7 at 15:19
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.