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

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- Sep 25th 2010, 04:25 PMrondo09Does the converse hold?
Suppose that both series $\displaystyle \Sigma a_k $ and $\displaystyle \Sigma b_k $ are absolutely convergent. Show that then so too is the series $\displaystyle \Sigma a_k b_k $. Does the converse hold?

- Sep 25th 2010, 08:30 PMHallsofIvy
Consider $\displaystyle a_k= b_k= \frac{(-1)^k}{k}$.

- Sep 25th 2010, 10:56 PMchisigma
- Sep 26th 2010, 02:43 AMHallsofIvy
Yes, the "converse" of "if A then B" is "if B then A".

- Sep 26th 2010, 05:58 AMchisigma
... very well!... what about $\displaystyle \displaystyle a_{k} = \frac{1}{k^{2}}$ and $\displaystyle \displaystyle b_{k} = \sin k$?... the series $\displaystyle \displaystyle \sum_{k=1}^{\infty} \frac{\sin k}{k^{2}}$ converges absolutely... is the same also for $\displaystyle \displaystyle \sum_{k=1}^{\infty} \sin k$ ?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Sep 26th 2010, 07:30 AMrondo09
- Sep 26th 2010, 07:39 AMchisigma
If $\displaystyle \Sigma a_{k}$ and $\displaystyle \Sigma b_{k}$ absolutely converge then is $\displaystyle \displaystyle \lim_{k \rightarrow \infty} |a_{k}| = \lim_{k \rightarrow \infty} |b_{k}| = 0$ , so that $\displaystyle \forall k>K$ is $\displaystyle \displaystyle |a_{k}\ b_{k}| < |a_{k}|$ or also $\displaystyle \displaystyle |a_{k}\ b_{k}| < |b_{k}|$ and the series $\displaystyle \Sigma a_{k}\ b_{k}$ converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$