# Convergent Sequence product Rule Proof.

• Mar 16th 2013, 01:45 AM
AlyoshaKaz
Convergent Sequence product Rule Proof.
Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l$ and $\displaystyle b_n - m$

$\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

Thanks.
• Mar 16th 2013, 08:33 AM
TheEmptySet
Re: Convergent Sequence product Rule Proof.
Quote:

Originally Posted by AlyoshaKaz
Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l$ and $\displaystyle b_n - m$

$\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

Thanks.

You just need to show that

$\displaystyle |a_nb_n-ml|< \epsilon$

So just add and substract $\displaystyle -a_nb+a_nb$ which is equal to zero to get

$\displaystyle |a_nb_n-mb_n+mb_n-ml|=|(a_n-m)b_n+(b_n-l)m|$

You still need to bound $\displaystyle b_n$ and finish this up.

Best wishes

IF you get stuck post back your workings and where your are stuck.
• Mar 16th 2013, 08:48 AM
Plato
Re: Convergent Sequence product Rule Proof.
Quote:

Originally Posted by AlyoshaKaz
Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l$ and $\displaystyle b_n - m$

$\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

If you carefully multiply the right side out, term by term, and combine then you will see how it all works.