- #1
Mathos
- 26
- 3
Homework Statement
Theorem. If lim a n = A and lim b n = B, then lim a n b n = AB.
Homework Equations
|an - A| < ε
|bn - 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 bn converges it is bounded and there exists an M1 such that for all n [itex]\in[/itex] N we have |bn| < M1. Define M1 [itex]\geq[/itex] 1. Then |bn| / M1 ≤ 1.
Skipping ahead, do the same for an and let |an - A| < [itex]\frac{ε}{2M1}[/itex]
and let
|bn - B| < [itex]\frac{ε}{2M2}[/itex].
|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|?