Theorem: If lim a n = A and lim b n = B, Then lim anbn = AB

[Mathos]

Homework Statement

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

Homework Equations

|a_n - A| < ε

|b_n - B| > ε

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 b_n converges it is bounded and there exists an M₁ such that for all n \in N we have |b_n| < M₁. Define M₁ \geq 1. Then |b_n| / M₁ ≤ 1.

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

and let

|b_n - B| < \frac{ε}{2M₂}.


|a_nb_n - AB| = |a_nb_n -Ab_n + Ab_n - AB|

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

So |a_nb_n -Ab_n + Ab_n - AB|

≤

|a_nb_n -Ab_n| + |Ab_n - AB|

=

|a_n - A||b_n| + |b_n - B||A|

<

(ε/2M₁)|b_n|

+

(ε/@M₂)|A|

Stupid question, but how are we multiplying each ε by |b_n| and |A|? I know that it gives us the two original lim a_n and lim b_n, 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|?

 

[SammyS]

Homework Statement

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

Homework Equations

|a_n - A| < ε

|b_n - B| > ε

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 b_n converges it is bounded and there exists an M₁ such that for all n \in N we have |b_n| < M₁. Define M₁ \geq 1. Then |b_n| / M₁ ≤ 1.

Skipping ahead, do the same for a_n and let |a_n - A| < \frac{\varepsilon}{2M_1}

and let

|b_n - B| < \frac{\varepsilon}{2M_2}.

|a_nb_n - AB| = |a_nb_n -Ab_n + Ab_n - AB|

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

So |a_nb_n -Ab_n + Ab_n - AB|

≤

|a_nb_n -Ab_n| + |Ab_n - AB|

=

|a_n - A||b_n| + |b_n - B||A|

<

(ε/2M₁)|b_n|

+

(ε/@M₂)|A|

Stupid question, but how are we multiplying each ε by |b_n| and |A|? I know that it gives us the two original lim a_n and lim b_n, 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|?

(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 a_n and B_n converge, so you can use any positive quantity to represent "ε" for each of these sequences.

 

[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.