Show the product of convergent sequences converge to the product of their limits

[k3k3]

Homework Statement

Use the fact that a_n=a+(a_n-a) and b_n=b+(b_n-b) to establish the equality

(a_n)(b_n)-ab=(a_n-a)(b_n)+b(a_n-a)+a(b_n-b)

Then use this equality to give a different proof of part (d) of theorem 2.7.

Homework Equations

The theorem it is citing is:
The sequence {(a_n)(b_n-b)} converges to ab.

The proof in the book uses bounds to establish that the theorem is true.

The Attempt at a Solution

Building (a_n)(b_n)-ab=(a_n-a)(b_n)+b(a_n-a)+a(b_n-b) is not hard, and I am omitting this part.

Assume a_n converges to a and b_n converges to b
Let ε>0 be given.
Using the definition of convergence, |a_n-a|< ε for all n≥ N_1 where N_1 is a positive integer.
Similarly, |b_n-b|< ε for all n≥ N_2 where N_2 is a positive integer.

Using the fact that (a_n)(b_n)-ab=(a_n-a)(b_n)+b(a_n-a)+a(b_n-b), then
|(a_n)(b_n)-ab|=|(a_n-a)(b_n)+b(a_n-a)+a(b_n-b)|

Using the triangular inequality and properties of absolute values, |(a_n)(b_n)-ab|=|(a_n-a)||(b_n)|+|b||(a_n-a)|+|a||(b_n-b)|

Then |(a_n-a)||(b_n)|+|b||(a_n-a)|+|a||(b_n-b)| < (ε)(ε)+|b|ε+|a|ε < ε

This is where my issues begin. My professor suggested using more N_x values. Four in total I believe. I need to show some how that the the product and sum is less then ε, but I am having trouble discovering how to set up the convergence definition to make the product and sum equal to ε.

 

[HallsofIvy]

a_nb_n- ab_n+ ab_n- ab= b_n(a_n- a)+ a(b_n- b)

 

[k3k3]

a_nb_n- ab_n+ ab_n- ab= b_n(a_n- a)+ a(b_n- b)

I am sorry but did you mean a_nb_n- ab_n+ ab_n- ab=b(a_n)+a(b_n-b) instead of b_n(a_n-a)?

Using that, then
|a_nb_n- ab_n+ ab_n- ab|=|b(a_n)+a(b_n-b)|<|b||a_n-a|+|a||b_n-b|

If |a_n-a|< ε/[2(|b|+1)] and |b_n-b|<ε/[2(|a|+1)], then

|b||a_n-a|+|a||b_n-b|< (ε|b|)/[2(|b|+1)]+(ε|a|)/[2(|a|+1)] = (ε/2)*[|b|/(|b|+1)]+(ε/2)*[|a|/(|a|+1)] < ε/2+ε/2=ε

Something like that?