# Thread: Does the converse hold?

1. ## Does 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?

2. Consider $\displaystyle a_k= b_k= \frac{(-1)^k}{k}$.

3. Originally Posted by rondo09
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?
Yust for clarity's sake: are You asking if, given a series $\displaystyle \Sigma a_{k} b_{k}$ that absolutely converges, the same holds for $\displaystyle \Sigma a_{k}$ and $\displaystyle \Sigma b_{k}$ ?...

Kind regards

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

4. Yes, the "converse" of "if A then B" is "if B then A".

5. ... 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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
What am I asking if there are series $\displaystyle \Sigma a_{k}$ and $\displaystyle \Sigma b_{k}$ which are absolutely convergent, then show that the series $\displaystyle \Sigma a_{k}b_{k}$ is also absolutely convergent. Then, P.S, does the converse hold?

7. 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

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