Transcript a n+1 - PPKE-ITK

Limits of Sequences of
Real Numbers
2013
Sequences of Real Numbers
Limits through Definitions
The Squeeze Theorem
Using the Squeeze Theorem
Monotonous Sequences
Index
FAQ
VIDEO and INTERNET SUPPORT
FOR THIS LECTURE
Explains the main points in THIS slide show:
http://www.youtube.com/watch?v=yBE1WApSpV4
Examples:
http://www.youtube.com/watch?v=hc64LUtPjP0
Theory through examples:

http://archives.math.utk.edu/visual.calculus/6/sequences.3/index.html
Index
FAQ
Sequences of Numbers
Definition
A sequence
 x1,x2 ,x3, 
is a rule that assigns,
to each natural number n, the number xn.
Examples
Index
1
 1 1 1
 1, 2 , 4 , 8 ,




2
1,1.4,1.41,1.414,1.4142, 
3
1, 3,5, 7,9, 
FAQ
Limits of Sequences
Definition
A finite number L is the limit of the sequence
 x1,x2 ,x3, 
if the numbers xn get arbitrarily close
to the number L as the index n grows.
If a sequence has a finite limit, then we say that
the sequence is convergent or that it converges.
Otherwise it diverges and is divergent.
Examples
Index
1
 1 1 1
The sequence 1, , , ,
 2 4 8
and its limit is 0.
0

 converges

FAQ
 1 1 1
The sequence 1, , , ,
 2 4 8
0
and its limit is 0.
Index

 converges

FAQ
Limits of Sequences
2
3
The sequence 1,1.4,1.41,1.414,1.4142,
and its limit is

converges
2.
The sequence (1,-2,3,-4,…) diverges.
Notation
lim xn  L
n 
Index
FAQ
Computing Limits of Sequences (1)
The limit of a sequence  xn  can be often computed by inserting n  
in the formula defining the general term xn . If this expression can be
evaluated and the result is finite, then this finite value is the limit of
the sequence. This usually requires a rewriting of the expression xn .
Index
FAQ
Computing Limits of Sequences
(1)
Examples
1
2
 1 1 1   1 
The limit of the sequence  1, , , ,    n 1  is 0 because
 2 4 8  2 
1
inserting n   to the formula xn  n 1 one gets 0.
2
 n2  1
The limit of the sequence  2 
 n  1
1
2
n2  1
n
is 1 because rewriting 2

n  1 1 1
n2
1
and inserting n   one gets 1.
1
n2
Index
0
FAQ
Computing Limits of Sequences
Examples continued
3
The limit of the sequence
n 1

n


n  1  n is 0 because of the rewriting
n 1 n

n 1 n

n 1 n
n  1  n

1


.
n 1 n
n 1 n
Insert n   to get the limit 0.
Index
FAQ
Formal Definition of Limits of
Sequences
Definition
A finite number L is the limit of the sequence
 x1,x2 ,x3, 
if
  0 : n such that n  n  L  xn   .
Example
1
 0 since if   0 is given, then
n  n
lim
1
1
1
0 
  if n   n .
n
n

Index
FAQ
Visualizing the formal definition
of a sequence

http://archives.math.utk.edu/visual.calcul
us/6/sequences.3/index.html
Index
FAQ
Immediate consequence of the
formal definition of a sequence
Theorem
Proof

Every convergent sequence is
bounded.
Suppose that lim xn=L . Take ϵ = 1 (any
number works). Find N 1 so that whenever
n > N1 we have xn within 1 of L. Then apart
from the finite set { a1, a2, ... , aN} all the terms
of the sequence are bounded by L+ 1 and L - 1.
So an upper bound for the sequence is max {x1
, x2 , ... , xN , L+ 1 }. Similarly one can find a
lower bound.
Index
FAQ
The Limit of a Sequence is
UNIQUE
Theorem
Proof

The limit of a sequence is UNIQUE
Indirectly, suppose, that a sequence would
have 2 limits, L1 and L2. Than for a given 
∃N 1 ∈N:∀n∈N:n>N 1 :|L1 −xn|<ϵ
∃N 2 ∈N:∀n∈N:n>N 2 :| L2 −xn|<ϵ
if N=max{N 1 ,N 2 }, xn would be arbitrary close
to L1 and arbitrary close to L2 at the same, it is
impossible-this is the contradiction (Unless L1
=L2)

Index
FAQ
Calculating limit using unique
prop.
Index
FAQ
Limit of Sums
Theorem
Assume that the limits lim xn  x and lim y n  y 
n 
are finite.
Proof
n 
Then lim  xn  y n   x  y  .
n 
Let   0 be given.
We have to find a number n with the property
n  n  xn  y n  x  y    .
To that end observe that also
Index

2
 0.
FAQ
Limit of Sums
Proof
Hence there are numbers n1 and n2 such that
n  n1  xn  x 


and n  n2  y n  y   .
2
2
Let now n =max  n1, n2  . We have
n  n  xn  y n  x  y   xn  x  y n  y  

2


2
 .
By the
Triangle
Inequality
Index
FAQ
Limits of Products
The same argument as for sums can be used to
prove the following result.
Assume that the limits lim xn  x and lim y n  y 
Theorem
n 
are finite.
Then
n 
lim xn y n  x y  .
n 
Remark
Observe that the limits lim xn y n and lim  xn  y n  may exist
n 
n 
and be finite even if the limits lim xn and lim y n do not exist.
n 
n 
1
. Then lim y n  0 and
2
n 
n
Examples
the limit lim xn does not exist. However, lim xn y n  0.
Let xn   1 n and y n 
n
n 
Index
n 
FAQ
Squeeze Theorem for Sequences
Theorem
Assume that n : xn  y n  zn and that
lim xn  lim zn  a.
n 
n 
Then the limit lim y n exists and
n 
lim y n  lim xn  lim zn .
n 
Proof
n 
n 
Let   0. Since lim xn  lim zn  a, nx  nz such
n 
n 
that n  nx  xn  a   and n  nz  zn  a   .
Let ny  max nx , nz . Then
n  ny  a  y n  max  a  xn , a  zn    .
This follows since xn  y n  znn.
Index
FAQ
Using the Squeeze/Pinching Theorem
Example
Solution
n!
.
n  nn
Compute lim
This is difficult to compute using the standard methods
because n! is defined only if n is a natural number.
So the values of the sequence in question are not given by an
elementary function to which we could apply tricks like L’Hospital’s
Rule.
n!
Here each term k/n < 1.
Observe that 0< n for all n  0.
n
Next observe that
n ! 1 2  3  n  1  n 1 2 3

  
n
n
nnn nn
n n n
n 1 n
1
  .
n n
n
Hence 0 
n! 1
 .
n
n
n
1
n!
 0, also lim n  0 by the Squeeze Theorem.
n  n
n  n
Since lim
Index
FAQ
Using the Squeeze Theorem
 sin(n ) 
Does the sequence 
 converge?
 n  cos(n ) 
If it does, find its limit.
Problem
Solution
 1  sin(n )  1
We have
Hence
and
 1  cos(n )  1 for all n  2,3,4, .
1
sin(n )
1



.
n  1 n  cos(n ) n  1
1
 1 
Since lim
 lim  
 0 we conclude that the sequence

n  n -1
n 
 n -1
 sin(n ) 
sin(n )
 0.

 converges and that nlim

n  cos(n )
 n  cos(n ) 
Index
FAQ
Monotonous Sequences
A sequence (a1,a2,a3,…) is increasing if an ≤ an+1 for all
n.
The sequence (a1,a2,a3,…) is decreasing if an+1 ≤ an for all n.
Definition
The sequence (a1,a2,a3,…) is monotonous if it is either increasing or
decreasing.
The sequence (a1,a2,a3,…) is bounded if there are numbers M and
m such that m ≤ an ≤ M for all n.
Theorem
A bounded monotonous sequence always has a
finite limit.
Observe that it suffices to show that the theorem for increasing
sequences (an) since if (an) is decreasing, then consider the
increasing sequence (-an).
Index
FAQ
Monotonous Sequences
A bounded monotonous sequence always has a finite limit.
Theorem
Let (a1,a2,a3,…) be an increasing bounded sequence.
Proof
Then the set {a1,a2,a3,…} is bounded from the above.
By the fact that the set of real numbers is complete,
s=sup {a1,a2,a3,…}
is finite.
Claim
lim an  s.
n 
Index
FAQ
Monotonous Sequences
A bounded monotonous sequence always has a finite limit.
Theorem
Let (a1,a2,a3,…) be an increasing bounded sequence.
Proof
Let s=sup {a1,a2,a3,…}.
Claim
lim an  s.
n 
Proof of the Claim
Let   0.
We have to find a number n with the property that n  n  an  s  .
Since s  sup an  , there is an element an such that s    an  s.
Since  an  is increasing n  n  s    an  an  s.
Hence n  n  an  s  . This means that lim an  s.
n 
Index
FAQ
SUMMARY




1. Notion of a sequence
2. Notion of a limit of a sequence
3. The limit of a convergent sequence is
unique.
4. Every convergent sequence is bounded.
5. Any bounded increasing (or decreasing)
sequence is convergent.
Note that if the sequence is increasing (resp. decreasing), then
the limit is the least-upper bound (resp. greatest-lower bound)
of the numbers
Index
FAQ
SUMMARY
6. If two sequences are convergent and we
compose their +, -, *. /, 1/.. then the limit
of this composed sequence exists and is
the +, -, *. /, 1/..of the original limiting
values.
7. Squeeze/Pinching theorem
Index
FAQ