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

In summary, the conversation discusses using 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) and using this equality to give a different proof of part (d) of theorem 2.7. The attempt at a solution involves using the definition of convergence and the triangular inequality to show that the product and sum are less than ε, using four different values of N_x.
  • #1
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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.
 
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  • #2
a_nb_n- ab_n+ ab_n- ab= b_n(a_n- a)+ a(b_n- b)
 
  • #3
HallsofIvy said:
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?
 

1. What is the definition of a convergent sequence?

A convergent sequence is a sequence of numbers that approaches a single fixed limit as the number of terms increases.

2. How do you determine if a sequence is convergent?

A sequence is convergent if the absolute value of the difference between consecutive terms approaches zero as the number of terms increases.

3. What is the product of convergent sequences?

The product of convergent sequences is the result of multiplying the terms of each sequence together.

4. What is the significance of the limit of a convergent sequence?

The limit of a convergent sequence represents the value that the sequence is approaching and is used to determine the convergence of the sequence.

5. How do you prove that the product of convergent sequences converge to the product of their limits?

To prove that the product of convergent sequences converge to the product of their limits, you would need to show that the absolute value of the difference between the product of the sequences and the product of their limits approaches zero as the number of terms increases. This can be done using the properties of limits and the definition of convergence.

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