# Intro to Real Analysis: Question on a completed proof.

• Nov 1st 2008, 08:57 AM
ilikedmath
Intro to Real Analysis: Question on a completed proof.
(The fact that http://www.cramster.com/Answer-Board...2187502633.gif is bounded is crucial.)

I found this:

http://i33.tinypic.com/2yxoxl2.jpg

My question is where the red arrow is pointing and the red boxed part:
How do I know how to choose 1/M for ε?
In this case we want to prove |$\displaystyle a_n$$\displaystyle b_n| < ε so I need to choose any ε value right? How do I know to choose 1/M? Is there algebra I can do that will lead me to choosing 1/M? I think I understand the rest of the proof up to that point where the red arrow is pointing to. Any help, suggestions are appreciated. Thanks for your time! • Nov 1st 2008, 09:15 AM Plato Quote: Originally Posted by ilikedmath How do I know how to choose 1/M for ε? In this case we want to prove |\displaystyle a_n$$\displaystyle b_n$| < ε
so I need to choose any ε value right? How do I know to choose 1/M?
Is there algebra I can do that will lead me to choosing 1/M?

You are to prove that given any $\displaystyle \varepsilon >0$ it is possible to insure that $\displaystyle \left| {a_n b_n - 0} \right| < \varepsilon$.
So from the fact that $\displaystyle \left( {a_n } \right) \to 0$ you use the definition of sequence convergence.
But in place of just $\displaystyle \varepsilon >0$ you are free to use $\displaystyle \frac{\varepsilon}{M} >0$.

$\displaystyle \left| {a_n b_n - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon$
• Nov 1st 2008, 10:21 AM
ilikedmath
Quote:

Originally Posted by Plato
You are to prove that given any $\displaystyle \varepsilon >0$ it is possible to insure that $\displaystyle \left| {a_n b_n - 0} \right| < \varepsilon$.
So from the fact that $\displaystyle \left( {a_n } \right) \to 0$ you use the definition of sequence convergence.
But in place of just $\displaystyle \varepsilon >0$ you are free to use $\displaystyle \frac{\varepsilon}{M} >0$.

$\displaystyle \left| {a_n b_n - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon$

So basically all they did was work with the inequality $\displaystyle \varepsilon >0$ and divided both sides by M to get the epsilon they wanted? :) It was that 'simple'? (Smirk)
• Nov 1st 2008, 10:27 AM
Plato
Quote:

Originally Posted by ilikedmath
So basically all they did was work with the inequality $\displaystyle \varepsilon >0$ and divided both sides by M to get the epsilon they wanted? :) It was that 'simple'?

NO that is not what happened.
$\displaystyle \frac{\varepsilon}{M} > 0$ was used in the definition of sequence convergence as the positive bound.
There are no "sides" here to be divided.
It is all a matter of properly using definitions.
• Nov 1st 2008, 12:43 PM
ilikedmath
Quote:

Originally Posted by Plato
It is all a matter of properly using definitions.

Ok, I see now. Thanks for clarifying.(Nod)