# Convergence proof...

Printable View

• Dec 20th 2011, 03:21 AM
breitling
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.
• Dec 20th 2011, 03:32 AM
girdav
Re: Convergence proof...
$0\leq |a_nb_n|\leq \frac{a_n^2+b_n^2}2$.
• Dec 20th 2011, 04:31 AM
chisigma
Re: Convergence proof...
Quote:

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 $c_{n}=\text{max}[|a_{n}|,|b_{n}|]$ it is clear that $\sum_{n} c^{2}_{n}$ converges. But because is $|a_{n}\ b_{n}|\le c^{2}_{n}$ also $\sum_{n} a_{n}\ b_{n}$ converges...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\chi$ $\sigma$
• Dec 20th 2011, 10:11 AM
breitling
Re: Convergence proof...
Many thanks to both of you