Real Analysis: Proving an*bn converges to ab

[TeenieBopper]

Homework Statement

Use the fact that an= a + (an - a) and bn= b + (bn - b) to establish the equality an*bn - ab = (an-a)(bn-b)+b(an-a)+a(bn-b).

Then, use this equality to prove that the sequence {an*bn} converges to ab.

Homework Equations

Definition of convergence: |an*bn - ab| < ε

The Attempt at a Solution

The first part was easy; just basic algebra. I'm stuck on the last part. I'm not sure where to begin. I tried expanding out the right side, hoping to find something I could use the triangle inequality on. I ended up with

an*bn - ab= an*bn -b*an -a*bn +ab+b*an-ab+a*bn-ab

= an*bn+a*bn-ab

= bn(an+a)-ab

I don't think I can do anything with that. Any suggestions where I can go from here? Am I even starting in the right place?

 

[Karamata]

Is the problem statement: if \displaystyle \lim_{n\to \infty} a_n = a and \displaystyle \lim_{n\to \infty} b_n = b proove that \displaystyle \lim_{n\to \infty} a_nb_n = ab??

 

[TeenieBopper]

Is the problem statement: if \displaystyle \lim_{n\to \infty} a_n = a and \displaystyle \lim_{n\to \infty} b_n = b proove that \displaystyle \lim_{n\to \infty} a_nb_n = ab??

Yes, sorry. Assume that an and bn converge to a and b, respectively.

 

[Office_Shredder]

All you need to do is show that (an-a)(bn-b)+b(an-a)+a(bn-b) goes to zero as n goes to infinity. If n is really big, what can you say about a_n-a? What about b_n-b? What does that say about the size of the entire expression?

The triangle inequality you might want to use is
|(a_n-a)(b_n-b)+b(a_n-a)+a(b_n-b)| \leq |(a_n-a)(b_n-b)|+|b(a_n-a)|+|a(b_n-b)|

 

[bizboy1]

Ya I agree with the above posters. This type of question is a triangle inequality question.