Cauchy product

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Melissa00

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Two series are given:
an with n=1 to infinity; an is absolute convergent.
bn is bounded (not specified though).

(1) Show that the series an*bn from n=1 to infinity is absolute convergent.
(2) If we change an with n=1 to infinity from absolute convergent to only being convergent, would the series an*bn from n=1 to infinity still be absolute convergent?

Am I right by saying that since bn is bounded, it converges?
For (1) I'd say that the product of the two series would be absolute convergent (because of the definition of the cauchy product) (?)
What about (2)? :confused:
 
Melissa00 said:
Two series are given:
an with n=1 to infinity; an is absolute convergent.
bn is bounded (not specified though).

(1) Show that the series an*bn from n=1 to infinity is absolute convergent.
(2) If we change an with n=1 to infinity from absolute convergent to only being convergent, would the series an*bn from n=1 to infinity still be absolute convergent?
Click to expand...

Suppose that \(\displaystyle B\) is the bound for \(\displaystyle b_n\),

(1) Do you know that \(\displaystyle \sum\limits_{n = 1}^\infty {\left| {{a_n}{b_n}} \right|} \le \sum\limits_{n = 1}^\infty {B\left| {{a_n}} \right|} \le B\left( {\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} } \right)~?\)


(2) Let \(\displaystyle {a_n} = \dfrac{{{{( - 1)}^n}}}{n}\,\& \,{b_n} = {( - 1)^{n }}\).

Does \(\displaystyle \sum\limits_{n = 1}^\infty {{a_n}{b_n}} \) converge?
 
pka said:
Explain your doubt!
Click to expand...

Okay, considering the cauchy product (1) converges?
Since in (1) an is absolute convergent and bn is bounded (converges as well), it fulfills the requirements of the cauchy product. Both series converge and at least one of them converges absolutely (in this case an).
Therefore their product converges as well. :confused:
 
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