# Thread: product of 2 series

1. ## product of 2 series

given that we have 2 series a_n and b_n that converges, does that mean that the product of this two series a_n b_n also converges?

what i said was that since both converge, the seq of finite partial sums for both will be bounded and hence
using cauchy seq,

sum from k = n to m of a_n b_n is also bounded.
since the product is cauchy, it is convergent.

is my reasoning right?

thanks

2. What if $\displaystyle a_n=b_n=\dfrac{(-1)^n}{\sqrt{n}}~?$

3. woops

4. of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.

5. Originally Posted by Krizalid
of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.
If $\displaystyle a_n\ge 0$ and $\displaystyle \sum {a_n }$ converges, then can you show that $\displaystyle \sum {(a_n)^2 }$ also converges?

Is it true that $\displaystyle 2a_nb_n\le (a_n)^2 +(b_n)^2~?$

6. Originally Posted by Krizalid
of course if the statement adds the assumptions $\displaystyle a_n,b_n>0$ then it would be true.
... it is sufficient that only $\displaystyle a_{n}$ or $\displaystyle b_{n}$ is $\displaystyle \ge 0$ and the other simply tends to 0...

Kind regards

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

7. Originally Posted by Plato
If $\displaystyle a_n\ge 0$ and $\displaystyle \sum {a_n }$ converges, then can you show that $\displaystyle \sum {(a_n)^2 }$ also converges?
The conditions imply that for $\displaystyle $$n large enough \displaystyle a_n<1, so \displaystyle a_n^2<a_n. So the sequence of partial sums of \displaystyle S_n=\sum_{n=0}^N {a_n^2 } is increasing and bounded by \displaystyle K+\sum_{n=0}^N {a_n } for some positive number \displaystyle$$K$, and so $\displaystyle S_N$ is increasing and bounded above ...

Is it true that $\displaystyle 2a_nb_n\le (a_n)^2 +(b_n)^2~?$