# Lim a(n)b(n) = AB

1. Feb 19, 2014

### Mathos

1. The problem statement, all variables and given/known data

Theorem. If lim a n = A and lim b n = B, then lim a n b n = AB.

2. Relevant equations

|an - A| < ε

|bn - B| > ε

3. The attempt at a solution

I have the solution to this, but I'm unclear on why one part is done.

Let ε > 0. Since bn converges it is bounded and there exists an M1 such that for all n $\in$ N we have |bn| < M1. Define M1 $\geq$ 1. Then |bn| / M1 ≤ 1.

Skipping ahead, do the same for an and let |an - A| < $\frac{ε}{2M1}$

and let

|bn - B| < $\frac{ε}{2M2}$.

|anbn - AB| = |anbn -Abn + Abn - AB|

So are we trying to manipulate the limit anbn into the familiar definition of |an - L| < ε so we can compare them?

So |anbn -Abn + Abn - AB|

|anbn -Abn| + |Abn - AB|

=

|an - A||bn| + |bn - B||A|

<

(ε/2M1)|bn|

+

(ε/@M2)|A|

Stupid question, but how are we multiplying each ε by |bn| and |A|? I know that it gives us the two original lim an and lim bn, but how can we just stick them on the other side of the "<" symbol? And how do |bn|/M1 = 1 and |A|/M2 = 1 if M is just defined as being larger than both? Can we just take the max of |bn| and |A|?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 19, 2014

### SammyS

Staff Emeritus
(In LaTeX, don't use the special characters at the right, like "ε" . Use \varepsilon . For subscripts use M_1, etc.)

First it will help to state what is required to show that $\displaystyle \lim\ \left(a_n\ b_n\right) = AB \$ .

You know that the sequences an and Bn converge, so you can use any positive quantity to represent "ε" for each of these sequences.

3. Feb 20, 2014

### dirk_mec1

|an - A||bn| + |bn - B||A|

You know that bn converges thus there is a M such that bn<M for all n. Since |an-A| can be made really small there is a n such that for all n> N there holds

|an - A| < epsilon/(2*(M+1)).

but then for all n>N:

|an - A||bn| < epsilon*M/(2*(M+1)) < epsilon /2.