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Limits of Sequences of
Real Numbers
Sequences of Real Numbers
Limits through Rigorous
Definitions
The Squeeze Theorem
Using the Squeeze Theorem
Monotonous Sequences
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
0
and its limit is 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,

converges
and its limit is 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
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 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