Download Math311W08Day3

Math 311 W08
Day 3
1. Read: 2.1 Page 25
2. Definition: A sequence is a real-valued function whose domain is the natural
numbers.
3. Notation: Rather than using f(n) notation for sequence, we use and denote the
entire sequence by {an}. And sometimes we just list out the elements of the
range, like {1, 2, 3, 4, 5, …}
4. Examples:
{1/n} also known as {1, 1/2, 1/3, 1/4, 1/5, …}
{1 + (-1)n} ***
also known as {0, 2, 0, 2, …}

if n  1
 2
an = 

 2  a n1 if n  1
***

also known as { 2 ,
2 2,
2 2 2 ,
2  2  2  2 , …}
5. Definition: The sequence {an} is said to converge if there exists a real number a
(called the limit of an) such that:
 


For every >0, there exists a natural number N such that nN  a n  a < 
(Picture on the board)
In this case, we often write an  a.

Examples:
{1/n} converges to 0. Proof: By the Archimedean Property, for any >0, there
exists a natural number N s.t. 1/N <. Thus if n N,
1
n
 0 = 1/n  1/N <. 
{1 + (-1)n} does not converge. Proof: Suppose the sequence converged to a 0.
a
 number N, there is an odd number
Then choose  = . Then given any natural
2
larger (e.g. 2N+1). So there is some n N such that 1  (1) n  a = a =2>.
Similarly, the sequence cannot converge to 0 since (with  =1) for any natural
numberthere is a larger even number (so there is some n N where the value of the
sequence is 2). Thus this term is not within epsilon of zero.
 So the sequence has no

limit and thus does not converge.
Now each of you try to prove the following: ***
n
5
A:
and
B:
sin(
) does not converge.

7

7
2
n2
Proof for A: Let >0. By the Archimedean Property we can choose a natural
5


 if n N, 2  7  7 = 5 1 2 < 5 1 <5 =.
 number N such that 1/N < . Then
n
n
5
5
n
Proof for B: First note that the sequence cannot go to zero. Choosing  = ½, note
that for any natural number, there is a bigger natural
form 4k + 1.
 number
 of the



For such an index the value of the sequence will be 1. Thus no matter how large
we choose our N, we will find terms of the sequence that are not within  ( ½) of 0.
a
. First note that 0 is not within
2
epsilon of a. Then for any natural number there is a larger natural number of the
form 4k. For such an index, the value of the sequence will be 0. Thus no matter
how large we choose our N, will find terms that are not within  of a.

Now, for any other real number a, we choose  =
Thus no number can be the limit of this sequence, so it does not converge. 
6. Comparison Lemma: Suppose an  a. Then the sequence {bn} converges to the
number b if there is a non-negative number C and an index N1 such that bn  b
 C a n  a for all n N1
Proof: Let >0. Let N2 be such that n N1 a n  a <


C 1
. 
Now let N = N1 + N2. Then if n N then bn  b  C a n  a < C

C 1
<.


7. Note: A lot of times people do not use the Comparison
Lemma, but instead use
the idea of the proof of the Comparison
Lemma
in other proofs. This was the


 been slightly shorter if
case with our example A above. Our proof would have
we had used the Comparison Lemma.
5
1
1
Proof: 1/n 0, and for all natural numbers n, 2  7  7 = 5 2  0 < 5 2  0 so
n
n
n
5
by the Comparison Lemma, 2  7  7. 
n
8. Definition: The sequence {an} is bounded if there exists some real number M



such that a n  M for every natural number n.
9. Theorem (Very
Useful): Convergent sequences are bounded. ***
 Suppose that an  a. Choose  = 1. There exists a natural number N such
Proof:
that n N  a n  a < 1. Then by a homework problem (sort of), we know that
a n < a + 1 for any n N. This gives us an upper bound for most of the guys in the
sequence. But we don’t know about the first N-1 terms. But there are only finitely
many
 of these things, so let’s just say M = Max { a1 , a 2 , …, a n1 , a + 1} and


this is our bound!

10. Theorem (The Bounded Away fromZero
that converge
 Theorem).
 Sequences

to nonzero elements are bounded away from zero in the following sense:
If an  a and a0. Let b be any positive real number such that b < a . Then there
exists a natural number N such that n N  a n > b. (In other words at some point
all remaining terms are more than b units away from zero.) ***


Proof: Suppose an  a and a0. Let b >0. Choose  = a -b. Then there is a natural
number N s.t. n N  a n  a < a -b. But then a < a n + a -b. Thus a n > b.
11.Theorem (Sum Property): Suppose that an  a and bn  b.
Then an + bn  a + b. ***



 



Proof: Let >0. Choose N1 such that n N1 a n  a < and choose N2 such
2

. Let N = max {N1 , N2}. Then n N
2
 
a n  bn  (a  b) = (a n  a)  (bn 
 b)  a n a + bn  b < + =. 
2 2
12.Theorem
(Product
Property):
Suppose
that
a

a
and
b

b.
n
n


Then an bn  ab.
that n N2 bn  b <




 
13. Questions about the proof:
a. Why did I choose a POSITIVE number M?
b. What is up with the “+1”?
c. Why does each step in the inequality chain work?
14. In the book the author proved a couple of cute little lemmas before proving the
Product Property Theorem. Since we did the theorem without the little lemmas
we can dismiss these with scorn and derision, beating them into submission
with our sledgehammer of a theorem!
Lemma 2.11: If an → a then for any number α, αan → αa.
Lemma 2.12: If an and bn both converge to zero, then so does an bn.
For practice, try to prove these directly (they are in the book so try them without
peeking). The book proves 2.11 using the Comparison Lemma. Try to also do it
with just the definition of convergence.
15. Theorem (Flippy Property). Suppose that the {bn} is a sequence of nonzero
1
1
 .
numbers that converge to the nonzero number b. Then
bn
b
Proof: (The book uses the Comparison Lemma, I will do it without).
b
away from zero theorem”,
 b by the “bounded

2
b
1 2
there is a natural number N1 such that n N1  bn >

<
bn b
2
that n N2  b  b <  b 2 .
Now choose N2 such
n
2


Then if we let N = Max{ N1 , N2 } then n N   


2  2
2 1
1 1 b  bn
1 1
b  bn < 2 b = .
=
 =
b  bn <
b b
bn b
bbn
bn b
b 2
Let >0. First note that since

16. Theorem (Quotient Property): If an → a and{bn} is a sequence of nonzero
 numbers
 that converge
  b. Then a n  a .
to 
the nonzero number
bn
b
Proof: This follows directly from the Flippy Property and the Product Property.
You could also prove it directly by using the Bounded Away from Zero Theorem
much like the previous proof (with the trick fromthe proof of the Product Property
thrown in).