1. ## Convergence proof...

I am a little stuck on the following:
suppose that:
1) sum(n..infinity) (a_n)^2 and sum(n..infinity)(b_n)^2 converges, then prove sum(n..infinity) (a_n)*(b_n) converges.

I am guessing it has something to do with the comparison test but i can't quite put my finger on it, something with 0<=(abs(a_n)-abs(b_n))^2

In addition am i on the right lines with thinking that, as the sum of (b_n)^2 converges, there must exist a number N such that sum(b_n)^2<1 for all n>=N
Then look at the sum that goes from N to infinity of a_n*b_n ? this is the part im stuck at,
any help would much appreciated, many thanks.

2. ## Re: Convergence proof...

$\displaystyle 0\leq |a_nb_n|\leq \frac{a_n^2+b_n^2}2$.

3. ## Re: Convergence proof...

Originally Posted by breitling
I am a little stuck on the following:
suppose that:
1) sum(n..infinity) (a_n)^2 and sum(n..infinity)(b_n)^2 converges, then prove sum(n..infinity) (a_n)*(b_n) converges.

I am guessing it has something to do with the comparison test but i can't quite put my finger on it, something with 0<=(abs(a_n)-abs(b_n))^2

In addition am i on the right lines with thinking that, as the sum of (b_n)^2 converges, there must exist a number N such that sum(b_n)^2<1 for all n>=N
Then look at the sum that goes from N to infinity of a_n*b_n ? this is the part im stuck at,
any help would much appreciated, many thanks.
Setting $\displaystyle c_{n}=\text{max}[|a_{n}|,|b_{n}|]$ it is clear that $\displaystyle \sum_{n} c^{2}_{n}$ converges. But because is $\displaystyle |a_{n}\ b_{n}|\le c^{2}_{n}$ also $\displaystyle \sum_{n} a_{n}\ b_{n}$ converges...

Marry Christmas from Serbia

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

4. ## Re: Convergence proof...

Many thanks to both of you