- #1
k3k3
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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.