# Does the converse hold?

• Sep 25th 2010, 04:25 PM
rondo09
Does the converse hold?
Suppose that both series $\Sigma a_k$ and $\Sigma b_k$ are absolutely convergent. Show that then so too is the series $\Sigma a_k b_k$. Does the converse hold?
• Sep 25th 2010, 08:30 PM
HallsofIvy
Consider $a_k= b_k= \frac{(-1)^k}{k}$.
• Sep 25th 2010, 10:56 PM
chisigma
Quote:

Originally Posted by rondo09
Suppose that both series $\Sigma a_k$ and $\Sigma b_k$ are absolutely convergent. Show that then so too is the series $\Sigma a_k b_k$. Does the converse hold?

Yust for clarity's sake: are You asking if, given a series $\Sigma a_{k} b_{k}$ that absolutely converges, the same holds for $\Sigma a_{k}$ and $\Sigma b_{k}$ ?...

Kind regards

$\chi$ $\sigma$
• Sep 26th 2010, 02:43 AM
HallsofIvy
Yes, the "converse" of "if A then B" is "if B then A".
• Sep 26th 2010, 05:58 AM
chisigma
... very well!... what about $\displaystyle a_{k} = \frac{1}{k^{2}}$ and $\displaystyle b_{k} = \sin k$?... the series $\displaystyle \sum_{k=1}^{\infty} \frac{\sin k}{k^{2}}$ converges absolutely... is the same also for $\displaystyle \sum_{k=1}^{\infty} \sin k$ ?...

Kind regards

$\chi$ $\sigma$
• Sep 26th 2010, 07:30 AM
rondo09
Quote:

Originally Posted by chisigma
Yust for clarity's sake: are You asking if, given a series $\Sigma a_{k} b_{k}$ that absolutely converges, the same holds for $\Sigma a_{k}$ and $\Sigma b_{k}$ ?...

Kind regards

$\chi$ $\sigma$

What am I asking if there are series $\Sigma a_{k}$ and $\Sigma b_{k}$ which are absolutely convergent, then show that the series $\Sigma a_{k}b_{k}$ is also absolutely convergent. Then, P.S, does the converse hold?
• Sep 26th 2010, 07:39 AM
chisigma
If $\Sigma a_{k}$ and $\Sigma b_{k}$ absolutely converge then is $\displaystyle \lim_{k \rightarrow \infty} |a_{k}| = \lim_{k \rightarrow \infty} |b_{k}| = 0$ , so that $\forall k>K$ is $\displaystyle |a_{k}\ b_{k}| < |a_{k}|$ or also $\displaystyle |a_{k}\ b_{k}| < |b_{k}|$ and the series $\Sigma a_{k}\ b_{k}$ converges...

Kind regards

$\chi$ $\sigma$