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?
... 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
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle \chi$ $\displaystyle \sigma$