# 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 $a_n=b_n=\dfrac{(-1)^n}{\sqrt{n}}~?$

3. woops

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

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

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

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

Kind regards

$\chi$ $\sigma$

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

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