Absolute Convergence

**hughmistler** · Oct 1st 2013, 11:26 AM

Let ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ be an absolutely convergent series of real numbers. Prove that the series ${\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}}$ converges.

Now by the definition of absolute convergence we know that:
${\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g}$ for some real number $g$ and ${\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h}$ for some real number $h$ .

So we have ${\sum\limits_{n = 1}^{\infty}{|(a_n)||(b_n)|}}={\sum\limits_{n = 1}^{\infty}{|(a_n)(b_n)|} = gh}$ .

How do I tie in square roots?

**Shakarri** · Oct 1st 2013, 01:11 PM

Originally Posted by hughmistler

[TEX]So we have ${\sum\limits_{n = 1}^{\infty}{|(a_n)||(b_n)|}}={\sum\limits_{n = 1}^{\infty}{|(a_n)(b_n)|} = gh}$ .

You cannot do this step because $\sum a_n \times \sum b_n \neq \sum a_nb_n$

$\sum a_n \times \sum b_n= (a_1+a_2+a_3...)(b_1+b_2+b_3+...)$

$\sum a_nb_n= a_1b_1+a_2b_2+a_3b_3+...$

Sorry I don't know how to go about doing the required proof.

**Plato** · Oct 1st 2013, 01:44 PM

Originally Posted by hughmistler

Let ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ be an absolutely convergent series of real numbers. Prove that the series ${\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}}$ converges.

For any real numbers $a~\&~b$ can you prove that $2\sqrt{|ab|}\le |a|+|b|~?$
Hint: look at $(\sqrt{|a|}-\sqrt{|b|})^2\ge 0.$

What can you do with that?

**bobcantor1983** · Oct 1st 2013, 07:58 PM

Originally Posted by Plato

For any real numbers $a~\&~b$ can you prove that $2\sqrt{|ab|}\le |a|+|b|~?$
Hint: look at $(\sqrt{|a|}-\sqrt{|b|})^2\ge 0.$

What can you do with that?

I mean, I understand how to derive that, i.e.,

$(\sqrt{|a|}-\sqrt{|b|})^2\ge 0.$
$(\sqrt{|a|}-\sqrt{|b|})(\sqrt{|a|}-\sqrt{|b|})\ge 0.$
$(|a| + |b| - 2\sqrt{|a||b|}\ge 0.$
$2\sqrt{|ab|}\le |a| + |b|.$

But I am not really sure how to use that. Do I need to use the definition of lim of sequence? (a number L is called the limit of a sequence if for any $\varepsilon > 0, \exists N \in \mathbb{N}$ such that for all $n > N$ , one has $|a_n - L| < \varepsilon$ .)

**Aryth** · Oct 1st 2013, 08:10 PM

Originally Posted by bobcantor1983

I mean, I understand how to derive that, i.e.,

$(\sqrt{|a|}-\sqrt{|b|})^2\ge 0.$
$(\sqrt{|a|}-\sqrt{|b|})(\sqrt{|a|}-\sqrt{|b|})\ge 0.$
$(|a| + |b| - 2\sqrt{|a||b|}\ge 0.$
$2\sqrt{|ab|}\le |a| + |b|.$

But I am not really sure how to use that. Do I need to use the definition of lim of sequence? (a number L is called the limit of a sequence if for any $\varepsilon > 0, \exists N \in \mathbb{N}$ such that for all $n > N$ , one has $|a_n - L| < \varepsilon$ .)

Note: $\frac{h + g}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)$

Note that the series on the right side is still convergent. What does this tell us about $\sum_{n=1}^{\infty}\sqrt{|a_n||b_n|}$ ?

**bobcantor1983** · Oct 1st 2013, 08:46 PM

I understand how you can do that with addition and that the sum of two convergent series converges, but how can that apply to multiplying two convergent series?

**Aryth** · Oct 1st 2013, 09:00 PM

Originally Posted by bobcantor1983

I understand how you can do that with addition and that the sum of two convergent series converges, but how can that apply to multiplying two convergent series?

By the absolute value property (and the how the square root is defined), we have that $\left(\frac{|a_n| + |b_n|}{2}\right)*\sqrt{|a_n||b_n|}\geq 0, \forall n$ . Following from my previous post, we have that:

$\sqrt{|a_n||b_n|} \leq \frac{|a_n| + |b_n|}{2}, \forall n$ .

By the comparison test, $\sum_{n=1}^{\infty}\sqrt{|a_n||b_n|}$ is convergent.

**bobcantor1983** · Oct 2nd 2013, 01:00 PM

So let me put this together, does this all make sense, am I missing anything?

Suppose ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ are absolutely convergent series of real numbers. We need to show that ${\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}}$ converges. We will show this using the comparison test.

The comparison test states: Suppose that we have two series, ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ , with $a_n, b_n \ge 0$ for all $n$ and $a_n \le b_n$ for all $n$ ; then if ${\sum\limits_{n = 1}^{\infty}{b_n}}$ is convergent so is ${\sum\limits_{n = 1}^{\infty}{a_n}}$ .

Now by the definition of absolute convergence we know that:
${\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g}$ for some real number $g$ and ${\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h}$ for some real number $h$ .

Let us consider that $\sqrt{|a_n b_n|} \le \frac{|a_n| + |b_n|}{2}$
We show this is true:

$(\sqrt{|a_n|}-\sqrt{|b_n|})^2\ge 0.$

$(\sqrt{|a_n|}-\sqrt{|b_n|})(\sqrt{|a_n|}-\sqrt{|b|})\ge 0.$

$(|a_n| + |b| - 2\sqrt{|a_n||b_n|}\ge 0.$

$2\sqrt{|a_n b_n|}\le |a_n| + |b_n|.$

$\sqrt{|a_n b_n|}\le \frac{|a_n| + |b_n|}{2}.$

Now consider that,
$\frac{g + h}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)$

Since we have shown that every element $\sqrt{|a_n b_n|}$ is less than or equal to this new series, and it is clear that this new series converges since it is the sum of two convergent series multiplied by a constant. Hence by the comparison test, $\sum_{n=1}^{\infty}\sqrt{|a_n||b_n|}$ is convergent.

**Plato** · Oct 2nd 2013, 03:24 PM

Originally Posted by bobcantor1983

So let me put this together, does this all make sense, am I missing anything?
Suppose ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ are absolutely convergent series of real numbers. We need to show that ${\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}}$ converges. We will show this using the comparison test.
The comparison test states: Suppose that we have two series, ${\sum\limits_{n = 1}^{\infty}{a_n}}$ and ${\sum\limits_{n = 1}^{\infty}{b_n}}$ , with $a_n, b_n \ge 0$ for all $n$ and $a_n \le b_n$ for all $n$ ; then if ${\sum\limits_{n = 1}^{\infty}{b_n}}$ is convergent so is ${\sum\limits_{n = 1}^{\infty}{a_n}}$ .
Now by the definition of absolute convergence we know that:
${\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g}$ for some real number $g$ and ${\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h}$ for some real number $h$ .
Let us consider that $\sqrt{|a_n b_n|} \le \frac{|a_n| + |b_n|}{2}$
We show this is true:
$(\sqrt{|a_n|}-\sqrt{|b_n|})^2\ge 0.$
$(\sqrt{|a_n|}-\sqrt{|b_n|})(\sqrt{|a_n|}-\sqrt{|b|})\ge 0.$
$(|a_n| + |b| - 2\sqrt{|a_n||b_n|}\ge 0.$
$2\sqrt{|a_n b_n|}\le |a_n| + |b_n|.$
$\sqrt{|a_n b_n|}\le \frac{|a_n| + |b_n|}{2}.$
Now consider that,
$\frac{g + h}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)$
Since we have shown that every element $\sqrt{|a_n b_n|}$ is less than or equal to this new series, and it is clear that this new series converges since it is the sum of two convergent series multiplied by a constant. Hence by the comparison test, $\sum_{n=1}^{\infty}\sqrt{|a_n||b_n|}$ is convergent.

There is nothing wrong with what you have done. It is just overly complicated.
From the given you know that both $\sum\limits_n {\left| {{a_n}} \right|} \;\& \;\sum\limits_n {\left| {{b_n}} \right|}$ converge.

From the lemma you know that
$\sum\limits_n {\left| {\sqrt {\left| {{a_n}{b_n}} \right|} } \right|} \le \sum\limits_n {\left| {2\sqrt {\left| {{a_n}{b_n}} \right|} } \right|} \le \sum\limits_n {\left| {{a_n}} \right|} \; + \;\sum\limits_n {\left| {{b_n}} \right|}$ .

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