Download n =N - Wku

Dr. Neal, WKU
MATH 337
Sequences
Let X be a metric space with distance function d . We shall define the general concept
of sequence and limit in a metric space, then apply the results in particular to some
special real-valued sequences.
∞
Definition 3.1. A sequence {x n }n =1 is an infinite list of points chosen from the metric
space X . Individual points xn are called the terms of the sequence. For a fixed integer
∞
N ≥ 1 , the terms {x n }n = N will be called the tail end of the sequence.
(A sequence also can be thought of as the list of function values attained for a function
f :ℵ→ X , where f (n) = xn for n ≥ 1.)
We want to make the notion of lim xn = x as precise as possible. This concept of
n →∞
∞
limit means that we can make all the tail end terms {x n }n = N as close as we like to the
point x ∈ X by choosing N large enough. Specifically, we can make d (xn , x) arbitrarily
small for all of the tail end terms provided N is sufficiently large enough. We formalize
the concept in the following definition.
Definition 3.2.
We say
lim xn = x , which may be written xn → x , provided the
n →∞
following condition holds: For every ε > 0 there exists an integer N ≥ 1 such that
d (xn , x) < ε whenever n ≥ N .
In this case, we say that {x n } converges to x and that x is the limit of {x n } . If no
such x ∈ X exists, then we say that {x n } does not converge in X or that {x n } diverges.
On the real line, lim xn = x provided: For every ε > 0 there exists an integer N ≥ 1
n →∞
such that x n − x < ε whenever n ≥ N . In any metric space, we have xn → x in X if
and only if d (xn , x) → 0 in ℜ .
x1
x N +k
x N +4
x5
x2
x N +3
xN
x4
x3
x
x N +2
x N +1
x6
Arbitrarily small ε - neighborhood about x
Eventually, all xn for n ≥ N (i.e., the terms x N , x N +1 , x N +2 , . . . ) are within ε of x .
Dr. Neal, WKU
n
= 1.
n →∞ n + 1
Example 3.1. On the real line, lim
Advanced Solution. Let ε > 0 be given. Let N be the smallest integer such that N ≥ 1
1
and N > − 1 . Then for n ≥ N , we have
ε
thus
n
→ 1.
n+1
n
n − (n + 1)
1
1
1
−1 =
=
≤
< 1
=ε;
n +1
n +1
n + 1 N + 1  − 1 + 1
ε

QED
The above proof is elegant and precise, but it does not indicate how N was found.
We shall now give an alternate proof that includes some more reasoning.
n ∞
1 2 3
=
, , , . . . is
n + 1 n=1
2 3 4
increasing and the terms are always less than 1. Thus the distance to 1 is
More Illustrated Solution.
We first note that the sequence
{ }
{
}
n
n
1
.
−1 = 1 −
=
n +1
n + 1 n +1
We wish to make this distance within an arbitrary ε > 0 by choosing n large enough.
1
1
n
1
So we need
< ε , which holds if < n + 1. Thus,
− 1 < ε provided n > − 1 .
n+1
ε
n +1
ε
The following result proves that 1 is the only possible limit of
{ n n+ 1 } .
Theorem 3.1. In a metric space, limits are unique. That is, if xn → x and xn → y , then
x = y.
Proof. To show that x = y , it suffices to show that d (x, y) < ε for all ε > 0 . So let ε > 0
be given. Because xn → x and xn → y , there exist integers N1 ≥ 1 and N2 ≥ 1 such that
d (xn , x) < ε / 2 for all n ≥ N1 and d (xn , y) < ε / 2 for all n ≥ N2 . Now choose any
integer n ≥ max( N1, N2 ) . Then
d (x, y) ≤ d( x, xn ) + d(x n , y) < ε / 2 + ε / 2 = ε .
{
Example 3.2. On the real line, 6 + (−1)
n ∞
diverges.
n= 1
}
QED
Dr. Neal, WKU
n
{
}
Solution. The terms of the sequence 6 + (−1) are 5, 7, 5, 7, 5, . . . which oscillate and
have no apparent limit. Also, successive terms are a distance of 2 apart. But suppose
6 + (−1)n → L for some L ∈ ℜ and let ε = 1 / 2 . Then there would be an integer N ≥ 1
n
such that 6 + (−1) − L < 1 / 2 whenever n ≥ N . But then for all n ≥ N , the triangle
inequality gives the distance between successive terms to be
(6 + (−1)n ) − (6 + (−1)n+1 ) = (6 + (−1)n ) − L + L − (6 + (−1) n+1 )
≤ 6 + (−1)n − L + L − (6 + (−1)n+1 )
< 1 / 2 +1 / 2 = 1 ,
which contradicts the fact that successive terms are a distance of 2 apart. Thus, no limit
n
L exists and 6 + (−1) diverges.
{
}
Definition 3.3. A sequence {x n } is bounded if there exists an x ∈ X and a real number
M > 0 such that d (xn , x) ≤ M for all n .
{
The sequence 6 + (−1)
are all within 1 of 6:
n
} also gives an example of a bounded sequence.
The terms
d (6 + (−1)n , 6) = 6 + (−1)n − 6 = (−1)n = 1 for all n .
n
{6 + (−1) } is a bounded sequence but it does not converge.
So
However all convergent
sequences are bounded as is proven next.
Theorem 3.2. If {x n } converges, then {x n } is bounded.
Proof. Assume xn → x . Then there exists an integer N ≥ 2 such that d ( xn , x ) ≤ 1 for all
n ≥ N . Now consider the terms x1 , . . . , x N −1 and the distances di = d ( x i , x ) for
Let M = max{d1 , . . ., dN −1, 1) .
Then d ( xn , x ) ≤ M for all n .
i = 1, . . ., N − 1.
QED
Corollary 3.1. If {x n } is not bounded, then {x n } diverges.
Corollary 3.2. Let {x n } be a sequence of real numbers. Then {x n } is bounded if and
only if there exists a real number c > 0 such that x n ≤ c for all n .
Proof. If such a c exists, then d (xn , 0) = xn ≤ c for all n ; thus, {x n } is bounded. Now
suppose {x n } is bounded with x ∈ ℜ and M > 0 such that x n − x = d( xn , x ) ≤ M for
all n . Now let c = M + x > 0 . Then for all n ,
xn = xn − x + x ≤ xn − x + x ≤ M + x = c .
Dr. Neal, WKU
{
Example 3.3. Consider the sequence 6 + (−1)
terms within 1 of 6. But in fact
n
} , which is bounded because it has all
6 + (−1) n = 6 + (−1)n − 6 + 6 ≤ 6 + (−1)n − 6 + 6 = 1 + 6 = 7 ; thus,
Thus, 6 + (−1)
n
≤ 7 for all n .
Theorem 3.3. In a metric space X , let x ∈ X and let E ⊆ X .
(a) The sequence {x n } converges to x if and only if every ε -neighborhood about x
contains all but a finite number of terms of {x n } .
(b) If x is a limit point of E , then there exists a sequence {x n } of distinct points in E
such that lim xn = x .
n →∞
Proof. (a) Suppose first that xn → x and let Nε (x ) be given. Then there exists an
integer N ≥ 1 such that d (xn , x ) < ε for all n ≥ N . That is, xn ∈ Nε (x ) for all terms
except possibly {x1, x2 , . . ., x N −1}.
Next, suppose that Nε (x ) contains all but a finite number of terms of {x n } for every
ε > 0 . So for every ε > 0 , there exists a finite set of terms Sε = {xn1 , xn2 , . . ., xnε } such
that xn ∈ Nε (x ) for all n ∉{n1 , . . ., nε } . Let N = max{n1 , . . ., nε } + 1. Then d (xn , x ) < ε for
all n ≥ N ; hence, xn → x .
(b) Because x is a limit point of E , every ε -neighborhood about x contains a point
from E that is different from x . Let ε1 = 1 . Then there exists x1 ∈ E, x1 ≠ x , such that
d (x1, x) < ε1 . Let ε 2 = min (1 / 2, d(x1 , x) ) . Then there exists x2 ∈ E, x 2 ≠ x , such that
d (x2 , x ) < ε 2 . Because d (x2 , x ) < d( x1 , x ) , we have x2 ≠ x1 . Continuing for n ≥ 3 , we let
ε n = min (1 / n, d(x n−1, x) ) and find an xn ∈ E, x n ≠ x, x n ∉{x1 , x2 , . . ., x n−1} , such that
d (xn , x ) < ε n . Then {x n } is a sequence of distinct points in E and we claim that
xn → x .
For if ε > 0 is given, we simply choose the smallest integer N such that 1 / N < ε .
Then for n ≥ N , we have d (xn , x ) < ε n ≤ 1 / n ≤ 1 / N < ε .
QED
The next results show how we can always find a sequence of rational numbers that
converge to a given irrational number.
Corollary 3.3. Let x be an irrational number. There exists a sequence {x n } of rational
numbers such that xn → x .
Proof. The rational numbers are dense on the real line; thus, every irrational number is
a limit point of the Rationals. The result follows from Theorem 3.3 (b).
Dr. Neal, WKU
Corollary 3.4. Let x = m. a1a2 a3 . . . be the decimal form of an irrational number. Let
xn = m. a1a2 a3 . . .an for n ≥ 1. Then xn are rational and xn → x .
Proof. Each xn is rational because it is written as a terminated decimal. Now let ε > 0
be given. Choose the smallest integer N such that 1 / 10 N < ε . Then for n ≥ N we have
x n − x = 0. 0 . . . an +1an+2 . . . ≤ 1 / 10n ≤ 1 / 10 N < ε . Thus, xn → x .
The next results are the common limit theorems from calculus:
Theorem 3.4.
Let {x n } and {y n } be real-valued sequences such that lim xn = x and
n →∞
lim yn = y . Then:
n →∞
(a) If xn = c for all n , then lim xn = c . (The limit of a constant is that constant.)
n →∞
(b)
lim c xn = c x , for all c ∈ℜ ; (Constants factor out of limits.)
n →∞
(c) lim (x n + yn ) = x + y ;
(The limit of a sum is the sum of the limits.)
(d)
(The limit of a product is the product of the limits.)
n →∞
lim (x n × yn ) = x y ;
n →∞
(e) lim 1 / xn = 1/ x , provided xn ≠ 0 for all n and x ≠ 0 .
n →∞
Proof.
(a) Let ε > 0 be given. For all n , x n − c = c − c = 0 < ε ; thus, xn → c .
(b) If c = 0 , then the result follows from (a). Otherwise, let ε > 0 be given. Then there
exists an integer N ≥ 1 such that x n − x < ε / c for all n ≥ N . Then for n ≥ N , we
have c xn − c x = c × x n − x < c × ε / c = ε ; thus, c xn → c x .
(c) Let ε > 0 be given. Because xn → x and yn → y , there exist integers N1 ≥ 1 and
N2 ≥ 1 such that x n − x < ε / 2 for all n ≥ N1 and yn − y < ε / 2 for all n ≥ N2 . Now let
N = max (N1, N2 ) . Then for n ≥ N , we have by the Triangle Inequality that
( xn + yn ) − (x + y ) = (x n − x) + (yn − y ) ≤ x n − x + yn − y < ε / 2 + ε / 2 = ε .
(d) Because {x n } converges, we know that {x n } is bounded. Thus, there exists M ∈ R ,
M > 0 such that xn ≤ M for all n . Because xn → x , we know by (b) that y xn → y x .
Now let ε > 0 be given.
There exist integers N1 ≥ 1 and N2 ≥ 1 such that
yn − y < ε / (2M ) for all n ≥ N1 and y x n − y x < ε / 2 for all n ≥ N2 . Now let
N = max (N1, N2 ) . Then for n ≥ N , we have by the Triangle Inequality that
Dr. Neal, WKU
x n yn − x y = x n yn − x n y + x n y − x y
≤ xn y n − xn y + x n y − x y
= x n yn − y + y x n − y x
≤ M yn − y + y xn − y x
ε
ε
<M
+
2M 2
=ε.
(e)
Because {x n } converges to x ≠ 0 , there exists an integer N1 ≥ 1 such that
x
for n ≥ N1 . Then for n ≥ N1 , we have
x − xn <
2
x
x = x − x n + x n ≤ x − xn + xn <
+ xn .
2
1
2
x
<
for n ≥ N1 .
< x n and then
xn
x
2
Let ε > 0 be given. There exists an integer N2 ≥ 1 such that x − x n < ε x 2 / 2 for
all n ≥ N2 . Now let N = max (N1, N2 ) . Then for n ≥ N , we have
By subtracting x / 2 , we have
1 1
x − xn
1
x − xn
2
ε x 2/2
−
=
=
×
<
×
= ε.
xn x
xn x
xn
x
x
x
Thus, 1 / xn → 1/ x .
QED
Monotone Real-Valued Sequences
We now consider the special case of real-valued sequences {x n } that are monotone in
the sense that either x1 ≤ x2 ≤ x3 ≤ . . .≤ x n . . . or x1 ≥ x2 ≥ x3 ≥ . . .≥ x n . . . .
Definition 3.4. Let {x n } be a sequence of real numbers.
(a) We say that {x n } is monotone increasing if xn ≤ x n+1 for all n . If xn < x n+1 for all n ,
then we say {x n } is strictly increasing.
(b) We say that {x n } is monotone decreasing if xn ≥ x n+1 for all n . If xn > x n+1 for all n ,
then we say {x n } is strictly decreasing.
∞
Theorem 3.5. Let {x n } be a sequence of real numbers such that {x n }n = K is monotone
increasing and bounded above for some K , or monotone decreasing and bounded
below for some K . Then there exists a real number x such that xn → x .
Dr. Neal, WKU
∞
Proof. Suppose {x n }n = K is monotone increasing and is bounded above. Then there
exists a real number M such that xn ≤ M for all n ≥ K .
That is,
x K ≤ x K +1 ≤ x K +2 ≤ . . .≤ M . By the Least Upper Bound Property of the Reals,
x = sup{ xn }∞
n = K exists and x ∈ ℜ . We assert that xn → x .
∞
Let ε > 0 be given. Then x − ε < x , so x − ε cannot be an upper bound of {x n }n = K
∞
because x is the least upper bound of {x n }n = K . So there exists an index N ≥ K such
that x − ε < x N ≤ x .
Then for all n ≥ N , we have x − ε < x N ≤ xn ≤ x ; thus,
x n − x = x − xn < x − ( x − ε ) = ε . Hence, xn → x .
We leave the proof for decreasing sequences that are bounded below as an exercise.
Proving that a Sequence is Monotone from Some Point
We now describe a formal method for proving that a sequence of real numbers is
monotone. If we want to show that x K ≤ x K +1 ≤ x K +2 ≤ . . . for some K , then we want
x
to prove that 1 ≤ n+1 for all n ≥ K . So we simply solve for the smallest index that
xn
satisfies the inequality. Alternately, we can solve the inequality xn+1 − xn ≥ 0 .
Likewise, if we want to show that x K ≥ x K +1 ≥ x K +2 ≥ . . . , then we solve for the
x
x
smallest index that satisfies the inequality 1 ≥ n+1 , or also written as n+1 ≤ 1.
xn
xn
Alternately, we can solve the inequality xn+1 − xn ≤ 0 .
Example 3.4. Consider the sequences below. Prove that each is strictly increasing or
strictly decreasing from some point. Then determine whether or not the sequence is
bounded. If the sequence is monotone and bounded, then find its limit.
(a)
n +1 ∞
n n=1
{ }

1 ∞
(b)  4 − 2 

n n =1
 n! ∞
(c)  n 
 6 n = 0
 n3 ∞
(d)  n 
 2 
n =0
n +1 ∞
are 2, 3/2, 4/3, 5/4, . . . which appear to be strictly
n n=1
n+1
decreasing from the onset. Claim: xn+1 < x n for n ≥ 1. Now let xn =
, and we shall
n
x
show that n+1 < 1. This ratio becomes
xn
n+2
2
xn +1
+ 1 = n + 2 × n = n + 2n < 1
= n
n+1
xn
n + 1 n + 1 n 2 + 2n + 1
n
for all n ≥ 1 because the numerator is less than the denominator and both are positive.
Thus, xn+1 < x n for all n ≥ 1, and the entire sequence is strictly decreasing.
Solution. (a) The terms of
{ }
Dr. Neal, WKU
n +1 ∞
are positive; thus, this sequence is clearly bounded
n n=1
below by 0. But in fact, 0 < x n ≤ x1 = 2 ; thus, xn ≤ 2 for all n ≥ 1. So this sequence is
bounded.
n+1
n+1
We now assert that
→ 1 (which also may be written as
↓ 1 ) . Let ε > 0 be
n
n
1
given. Let N be the smallest integer such that N > . Then for n ≥ N we have
ε
All the terms of
{ }
n +1
1 1
−1 = ≤ < ε .
n
n N
------------------------------------------------------------------
1 ∞
(b) The terms of  4 − 2 
are (4 – 1), (4 – 1/4), (4 – 1/9), (4 – 1/16), . . . which

n n =1
appear to be strictly increasing from the onset and nearing 4. Claim: xn < x n+1 for all
n ≥ 1. It is clear that 3 ≤ x n < 4 for all n ≥ 1; thus, the sequence is bounded.
x
1
Now let xn = 4 − 2 , and we shall try to show that n+1 > 1 for n ≥ 1. In this case,
xn
n
we shall instead show that xn+1 > x n or that xn+1 − xn > 0 . We then have

1  
1
1
1
 −4 − 2 = 2 −
xn+1 − xn =  4 −
> 0 for n ≥ 1
2

(n + 1)  
n  n
(n + 1)2
because 0 < n2 < (n + 1)2 which makes 1 / n 2 > 1 / (n + 1) 2 .
1
1
We now assert that 4 − 2 → 4 . To prove this, we shall show instead that 2 → 0
n
n
and then apply limit theorems. Let ε > 0 be given. Let N be the smallest integer such
1
1
2
that N >
. Then N > . Then for n ≥ N we have
ε
ε
1
1
1
2 − 0 = 2 ≤ 2 < ε.
n
n
N

1
1 
1
Thus, 2 → 0 ; hence, lim  4 − 2  = lim 4 + (−1) lim 2 = 4 + (−1)0 = 4.


n →∞
n
n →∞
n→∞ n
n
------------------------------------------------------------------ n! ∞
(c) We claim that the sequence  n 
becomes strictly increasing after a certain
 6 n = 0
n!
point. Let xn = n . Then consider the inequality
6
(n + 1)!
n+1
x
(n + 1)! 6 n n + 1
1 < n+1 = 6 n! = n+1 ×
=
.
xn
n!
6
6
6n
Dr. Neal, WKU
This inequality holds if and only if 6 < n + 1 if and only if 5 < n . Thus for n ≥ 6 , we
have xn < x n+1 . The first few terms of this sequence are x0 = 1 , x1 = 1 / 6 , x2 = 2 / 36 ,
x3 = 6 / 216 , x4 = 24 / 1296 , x5 = 120 / 7776 , x6 = 720 / 46656 , x7 = 5040 / 279936 , . . .
Note that x0 > x1 > x 2 > x3 > x4 > x 5 = x6 , but then x6 < x 7 < x 8 < . . . .
 n! ∞
Next, we assert that  n 
is not bounded. Suppose this sequence were
 6 n = 0
bounded by M . Then note that 1 < x14 ≤ xn for n ≥ 14 . Now adjust the inequality
x
n +1
above: M < n+1 =
, which is satisfied for 6M − 1 < n . Then for n ≥ max(6M, 14) ,
xn
6
x
because xn > 1 we have xn +1 > n +1 > M , which contradicts the fact that M bounds the
xn
sequence. Hence the sequence is not bounded and therefore must diverge.
------------------------------------------------------------------∞
 n3 


(d) We claim that the sequence
becomes strictly decreasing after a certain
 2n 
n =0
n3
point. Let xn = n . Then consider the inequality
2
(n + 1)3
n+1
x n+1
(n + 1)3 n 3 + 3n2 + 3n + 1
2
.
1>
=
=
=
xn
n3
2n 3
2n3
2n
3
3
2
This inequality holds if and only if 2n > n + 3n + 3n + 1 if and only if
3
n − 3n2 − 3n − 1 > 0 . Because the cubic polynomial n3 − 3n2 − 3n − 1 has a positive
leading coefficient, it grows without bound; thus, there is an integer N such that
3
2
n − 3n − 3n − 1 > 0 for all n ≥ N . We simply need to find the smallest such integer N .
3
2
By trial and error, we see that for n ≥ 4 , we have n − 3n − 3n − 1 > 0 and thus
xn > x n+1 .
∞
 n3 


We now know that
is strictly decreasing. All the terms are positive; thus
 2n 
n =4
we have 0 < x n ≤ x4 = 4 for all n ≥ 4 . In fact, 0 < x n ≤ 4 for n = 0, 1, 2, 3 also; thus, the
 n3 ∞
original sequence is bounded. Because  n 
is strictly decreasing and bounded
 2 
n =4
below by 0, it must have a limit. In fact
Rule which we shall prove later.
n3
2n
→ 0 . This result follows from L'Hopital‘s
Dr. Neal, WKU
Divergent Real-Valued Sequences
If {x n } does not converge, then it diverges. One type of divergence is an “oscillation”
for which the sequence remains bounded but varies back and forth between several
n ∞
terms and never attains a limit. For example, consider {(−1) }n= 0 . Then the terms are
1, –1, 1, –1, 1, –1, . . . , which never have a limit. Another example is { sin ( n π / 4) }∞
n =0,
which has terms 0, 2 / 2 , 1, 2 / 2 , 0, – 2 / 2 , –1, – 2 / 2 , 0, 2 / 2 , 1, . . .
Another type of divergence occurs when a sequence is unbounded. (Recall: A
convergent sequence must be bounded.) When terms tend to get larger and larger, or
tend to get smaller and smaller, they may be said to have an “infinite limit.”
Definition 3.5. Let {x n } be a sequence of real numbers.
(a) We say that
lim xn = + ∞ provided for every positive real number M , there exists
n →∞
an integer N ≥ 1 such that xn > M whenever n ≥ N .
(b) We say that
lim xn = − ∞ provided for every negative real number M , there exists
n →∞
an integer N ≥ 1 such that xn < M whenever n ≥ N .
Example 3.6. (a) Prove that the following limits diverge to either +∞ or –∞.
n ∞
3 ∞
(i) {2 }n= 0
(ii) {−n }n= 0
n 2 ∞
diverges but does not have an infinite limit.
n= 0
{
(b) Explain why (−1) n
}
n
Solution. (a) (i) Let M > 0 . We want to make 2 > M , so we need n > ln M / ln 2 . Let N
n
be the smallest integer such that N > ln M / ln 2 . Then 2 > M for all n ≥ N . We note
n
that {2 n} is in fact a strictly increasing sequence; thus, 2 ↑ ∞ .
3
3
(ii) Let M < 0 . We want to make − n < M , so we need n > − M . Let N be the smallest
3
3
integer such that N > − M . Then − n < M for all n ≥ N . We note that {−n3 } is in fact
3
a strictly decreasing sequence; thus, − n ↓ −∞ .
(b) The terms of (−1)n n2
{
∞
{(−1)2 k (2k )2 }k= 0
=
∞
}n=0 are 0, –1, 4, –9, 16, –25, . . .
∞
{(2k )2 }k =0 = {0, 4, 16, . . . } that is
There is a subsequence
unbounded; thus, this
subsequence diverges. The original sequence must also diverge, because if it
converged, then every subsequence would also converge (Theorem 3.9 (a) to come).
Because the terms of the sequence oscillate from positive to negative, there is no
way to make xn > 1000 for all n ≥ N no matter the N . Thus, the sequence cannot have a
limit of +∞. Likewise, the limit cannot be –∞.
Dr. Neal, WKU
Theorem 3.6. If lim xn = + ∞ or if lim xn = − ∞ , then lim 1 / xn = 0 .
n →∞
n →∞
n →∞
Proof. Assume lim xn = + ∞ . Let ε > 0 be given. There exist an integer N ≥ 1 such that
n →∞
xn > 1 / ε > 0 whenever n ≥ N . But then ε > 1 / xn > 0 or 1 / x n − 0 < ε for all n ≥ N ;
thus, 1 / xn → 0 . The proof is similar if lim xn = − ∞ .
n →∞
Some Special Real-Valued Sequences
Here are the limits of a few common real-valued sequences:
p
p
(i) For p > 0 , lim n = + ∞ and lim 1 / n = 0 .
n →∞
n →∞
(ii) lim n! = +∞ and lim 1 / n! = 0
n →∞
n →∞
n
(iii)
n
(a) If −1 < x < 1 , then x → 0
(b) If x = 1 , then x → 1
n
n
(c) If x > 1 , then x → +∞
(iv) For x > 0 , lim
n →∞
(v)
n
(d) If x ≤ −1, then x diverges (oscillates)
x = 1.
lim n n = 1
n →∞
Rationale
p
1/ p
(i) For any M > 0 , we know n > M for n > M
and (n + 1) p > n p for all n ≥ 1; thus
p
n is unbounded and strictly increasing for n ≥ 1. Hence, n p ↑ ∞ and 1 / n p ↓ 0 .
{ }
------------------------------------------------------------------(ii) We know n! = n × (n − 1) × . . . × 1 ≥ n > 0 and (n + 1)! = (n + 1) × n! > n! for all n ≥ 1;
thus { n! } is unbounded and strictly increasing for n ≥ 1. Hence, n! ↑ ∞ and 1 / n! ↓ 0 .
------------------------------------------------------------------n
n
(iii) (a) Let −1 < x < 1 . If x = 0 , then x = 0 for all n ≥ 1; thus, x → 0 . Now suppose
x ≠ 0 , so that 0 < x < 1 and ln x < 0 .
Then let ε > 0 be given.
We want
n
n
n
x − 0 = x = x < ε for n large enough. We simply need n > ln ε / ln x . Thus,
n
x → 0.
Dr. Neal, WKU
n
n
(b) If x = 1 , then x = 1 for all n ≥ 0 ; thus, x → 1 .
n
(c) Suppose x > 1 . Then for any M > 0 , we know x > M for n > ln M / ln x and
n+1
= x × x n > x n for all n ≥ 1; thus {x n } is unbounded and
because x > 1 , we have x
strictly increasing for n ≥ 1. Hence, x n ↑ ∞ (and 1 / x n ↓ 0 ).
(d) For x ≤ −1, the terms of {x n } oscillate between positive and negative for even and
n
odd exponents of n . Moreover, the terms x = x n are also unbounded for x < −1 (they
are bounded though when x = −1 ). So it is clear that {x n } is oscillatingly divergent.
------------------------------------------------------------------n
1/n
(iv) Assume x > 0 , and consider x = x . We take the natural log of this expression
1
1
to obtain ln x . Because → 0 by sequence (i) above and ln x is a constant, we have
n
n
1
n
0
ln x → 0 . Thus, x → e = 1 . (Note: This argument uses a result about continuous
n
functions to be proven later: If ln( xn ) → x , then xn → e x .)
------------------------------------------------------------------ln n
n
(v) To show lim n = 1, we also need the fact that
→ 0 (which can be obtained
n
n →∞
1
n
0
n
from L’Hopital‘s Rule to come later). Then ln n = ln n → 0 ; hence, n → e = 1 .
n
------------------------------------------------------------------Cauchy Sequences
A special type of sequence is one in which the terms become closer and closer together.
Our next main goal is to show that such sequences in ℜ will always have a limit.
Definition 3.5. Let {x n } be a sequence in a metric space X with metric d . Then {x n } is
called a Cauchy sequence provided for every ε > 0 , there exists an integer N ≥ 1 such that
d ( xn , xm ) < ε whenever n, m ≥ N .
x1
A Cauchy Sequence
x3
eventually all close together
x2
→
x N +1
x N x N +2
x N +k
Dr. Neal, WKU
Theorem 3.7. If {x n } converges, then {x n } is Cauchy.
Proof.
Let xn → x . Then there exists an integer N ≥ 1 such that d ( xn , x ) < ε / 2
whenever n ≥ N . Then for all n, m ≥ N , we have
d ( xn , xm ) ≤ d ( xn , x ) + d ( x, x m ) < ε / 2 + ε / 2 = ε .
QED
Theorem 3.8. A Cauchy sequence is bounded.
Proof.
d ( xn ,
d ( xn ,
QED
Let {x n } be Cauchy, and choose the smallest integer N ≥ 2 such that
xm ) < 1 whenever n, m ≥ N . Let M = max{d ( x1 , x N ), . . . ,d ( x N −1 , x N ), 1} . Then
x N ) ≤ M , so all points of {x n } are within M of x N . Thus, {x n } is bounded.
Corollary 3.5. (Completeness of the Reals) Let {x n } be a Cauchy sequence of real
numbers. There exists x ∈ℜ such that xn → x .
Proof. Let E = {x n } be the set of distinct terms in the sequence. Suppose first that E is
finite with S = { x1 , xn2 , . . . , xnk } being the finite set of distinct terms labeled with
increasing indices. Then there are only a finite number of pairs of distinct elements that
can be chosen from S . So let ε = min{d (xi , x j ) : x i , x j ∈ S, xi ≠ x j } . Because the
sequence is Cauchy, we can choose an integer N > nk such that d ( xn , xm ) < ε / 2 for all
n, m ≥ N . But x N must equal one of the elements in S as must xn for n ≥ N . Because
d ( xn , x N ) < ε for n ≥ N , we must have xn = x N for all n ≥ N . Hence, xn → x N ∈ ℜ .
Next, suppose that E = {x n } is infinite. Because {x n } is Cauchy, E is bounded. So
by the Bolzano-Weierstrass Theorem, there exists x ∈ℜ such that x is a limit point of
E . We claim that xn → x .
Let ε > 0 be given. There exists an integer N ≥ 1 such that xn − xm < ε / 2 for all
n, m ≥ N . We now want to choose a smaller neighborhood around x . Let ε1 > 0 such
that ε1 < ε / 2 and ε1 < min{ x − x i : 1 ≤ i < N, x ≠ xi }. Because x is a limit point of E ,
there exists a point x K from E (different from x ) such that x − x K < ε 1. By the choice
of ε1 , we must have K ≥ N . Then for all n ≥ N ,
x − xn = x − xK + x K − xn ≤ x − x K + xK − xn < ε1 + ε / 2 < ε
Hence, xn → x .
QED
Thus, on the real line ℜ , every Cauchy sequence converges. Any metric space that
has this property is said to be complete. Thus, ℜ is complete.
Dr. Neal, WKU
Subsequences
Let {x n } be a sequence in a metric space. The terms are indexed as x1 , x2 , x3 , . . , xn . . .
Suppose we want to choose a partial infinite list of these, such as x3 , x6 , x9 , . . . , x3k . . .
Such a partial list is called a subsequence and is denoted by x nk . The terms of the
subsequence are xn1 , xn2 , xn3 , . . . , xnk . . .
So n1 is the smallest index chosen from the original sequence that puts xn1 in the
subsequence. So n1 ≥ 1. Then n2 is the next smallest index chosen from the original
sequence that puts xn2 in the subsequence. Thus n2 ≥ 2 (and n2 > n1 ). In general, we
always must have nk ≥ k and ( nk +1 > nk ).
{ }
Example 3.7.
Then
x1 = 12
Let xn = n2 for n ≥ 1. Let xnk be the square of the k th prime number.
x2 = 22
x3 = 32
xn1 = 22
xn2 = 32
So, n1 = 2 ≥ 1
x4 = 4 2
n2 = 3 ≥ 2
x5 = 52
x6 = 62
xn3 = 52
n3 = 5 ≥ 3
x7 = 72
x8 = 82 . . .
xn4 = 72
...
nk = kth prime ≥ k
Theorem 3.9. Let {x n } be a sequence in a metric space. (a) If lim xn = x , then for
n →∞
{ }
every subsequence x nk , we have lim xnk = x .
k→ ∞
{ }
(b) If {x n } is Cauchy and lim xnk = x for some subsequence x nk , then lim xn = x .
k→ ∞
n →∞
Proof. (a) Let ε > 0 be given. There exists an integer N ≥ 1 such that d (xn , x) < ε for all
n ≥ N . The N th index n N of a subsequence must be at least N ; that is, n N ≥ N . So for
k ≥ N , we have nk ≥ n N ≥ N and therefore d (xnk , x ) < ε . Hence, lim xnk = x .
k→ ∞
(b) Let ε > 0 be given. There exists an integer K1 ≥ 1 such that d (xnk , x ) < ε / 2 for all
k ≥ K1 . There also exists an integer K2 such that d (xn , xm ) < ε / 2 for all n, m ≥ K2 .
Now let N = max (K1 , K2 ) and let n ≥ N . Then n N ≥ N ≥ K2 and N ≥ K1 , thus
d (xn , x) ≤ d (x n , xn N ) + d(x nN , x ) < ε .
QED
Dr. Neal, WKU
Note: Just because a subsequence converges, it does not mean that the original nonCauchy sequence converges. For example, let xn = 6 + (−1)n for n ≥ 1: 5, 7, 5, 7, 5, 7, . . .
Then let xnk = x 2k = 6 + (−1)
diverges.
2k
for k ≥ 1 : 7, 7, 7, 7, . . .
Then xnk → 7 but {x n }
Theorem 3.10. Let x be a limit point of a sequence {x n } in a metric space. Then there
exists a subsequence x nk such that lim xnk = x .
{ }
k→ ∞
Proof. Because x is a limit point of a {x n } , every ε -neighborhood about x contains
infinitely many points from {x n } . Let ε1 = 1 / 2 and let n1 be the first index such that
xn1 ∈ Nε1 (x ) . Next let ε 2 = 1 / 4 and choose the first index n2 > n1 such that
xn2 ∈ Nε 2 (x ) . Continuing, let ε k +1 = 1 / 2 k+1 and choose the first index nk +1 > nk such
k
that xnk+1 ∈ Nε k+1 ( x) . Then d (x, x nk ) < ε k = 1 / 2 for k ≥ 1 , which implies lim xnk = x .
k→ ∞
QED
Corollary 3.6. In ℜ , every bounded sequence has a convergent subsequence.
Proof. Suppose first that there are only a finite number of distinct terms in {x n } . Then
at least one term must recur an infinite number of times in the list x1 , x2 , x3 , . . . We let
x denote the value of this recurring term, then choose n1 < n2 < n3 < . . . such that
xnk = x for all k ≥ 1 . Then lim xnk = x .
k→ ∞
Next suppose that there are an infinite number of terms in {x n } . Then the sequence
creates an infinite, bounded set of real numbers. By the Bolzano-Weierstrass Theorem,
this set has a limit point x . The result now follows from Theorem 3.10.
QED
Lim sup and Lim inf
Let {x n } be a sequence of real numbers. Then {x n } may or may not converge, but some
subsequences may converge. If {x n } has some convergent subsequences, then let E be
the set of all limits of all convergent subsequences of {x n } including those limits that
are ± ∞ . Then E may or may not be bounded. If E has no lower bound, then let
inf E = –∞ . If E has no upper bound, then let sup E = + ∞ . If bounds exist, then inf E
and sup E exist as real numbers by the glb and lub properties of ℜ .
Definition 3.7. Let {x n } be a sequence of real numbers and let E be the set of all limits
of all convergent subsequences of {x n } . We define the lim inf and lim sup of {x n } by
x∗ = lim inf x n = inf E
n→ ∞
and
x ∗ = lim sup x n = sup E .
n→ ∞
Dr. Neal, WKU
Example 3.8. (a) Let {x n } be the set of all rational numbers. Every irrational number x
is a limit point of the Rationals, thus there exists a subsequence of the Rationals that
converges to x . So the set E of all limits of all convergent subsequences of {x n } is not
bounded; thus, x∗ = inf E = − ∞ and x ∗ = sup E = + ∞ .

2
2 
, 0,
, 1 . Thus, x∗ = inf E = −1
(b) Let xn = sin( n π / 4 ) for n ≥ 0 . Then E =  −1, −


2
2
x ∗ = sup E = 1
We next give a result that relates a limit with the lim inf and lim sup:
Theorem 3.11. Let {x n } be a sequence of real numbers and let x ∈ℜ . Then lim xn = x
n →∞
if and only if lim inf x n = x = lim sup xn .
n→∞
Proof.
Assume
n→ ∞
lim xn = x , and let E be the set of all limits of all convergent
n →∞
subsequences of {x n } , including possibly ± ∞ . By Theorem 3.9 (a), E = { x }; thus,
lim inf x n = inf E = x = sup E = lim sup xn .
n→∞
n→ ∞
Next, assume that lim inf x n = x = lim sup xn . We must show that xn → x . We first
n→∞
n→ ∞
assert that {x n } is bounded. For if {x n } were not bounded above, then there would be
a subsequence x nk that increases to +∞ and then lim sup xn = sup E = + ∞ ∉ R . So
{ }
n→ ∞
{x n } must be bounded above, and likewise must be bounded below.
Next, let E be the set of all limits of all convergent subsequences of {x n } . We assert
that E = { x }. For if there were two points x1 , x 2 ∈ E with x1 < x2 , then we would have
lim inf x n = inf E ≤ x1 < x2 ≤ sup E = lim sup xn . Thus, E = { x }.
n→∞
n→∞
Now suppose xn →
/ x . Then there exists an ε > 0 such that for every integer N ≥ 1
there exists an n ≥ N such that xn − x ≥ ε . For this ε , choose the smallest integer n1 ≥ 1
such that xn1 − x ≥ ε . Having chosen nk , let nk +1 be the smallest integer with
{ }
nk +1 ≥ nk + 1 such that xnk+1 − x ≥ ε . We then obtain a subsequence x nk with all
terms in (− ∞, x − ε ] ∪ [x + ε , ∞) . At least one of these rays must have an infinite
number of terms of x nk . If [ x + ε , ∞ ) has an infinite number of the terms, then these
terms in [ x + ε , ∞ ) are an infinite, bounded set. By Bolzano-Weierstrass, there must be
a real-valued limit point y to this set, and by Theorem 3.10, there must be a further
subsequence of the terms that converges to y . Thus, y ∈ E = {x}. But because all the
terms are in [ x + ε , ∞ ) , we must have x + ε ≤ y , which is a contradiction. Likewise, we
obtain a contradiction if (− ∞, x − ε ] has an infinite number of terms. Hence, we must
have xn → x .
QED
{ }
Dr. Neal, WKU
Exercises
1. Let {x n } and {y n } be real-valued sequences such that lim xn = x and lim yn = y
n →∞
n →∞
where xn ≠ 0 for all n and x ≠ 0 . Prove that lim yn / x n = y / x .
n →∞
∞
2. Let {x n } be a sequence of real numbers such that {x n }n = K is monotone decreasing
and bounded below for some K . Prove that there exists an x ∈ ℜ such that xn → x .
3. Let {x n } be a sequence of real numbers.
(a) Prove that if {x n } converges, then { xn } converges.
(b) If { xn } converges, does it necessarily follow that {x n } converges?
disprove.
(c) Prove that if { xn } converges to 0, then {x n } converges to 0.
Prove or
4. Let {x n } be a sequence of real numbers that is bounded above by M and such that
xn → x . Prove that x ≤ M .
5. Let {x n } and {y n } be real-valued sequences.
(a) Suppose xn → 0 and {y n } is bounded. Prove that xn yn → 0 .
(b) Suppose xn → + ∞ and {y n } is bounded. Prove that yn / xn → 0 .
6. In a metric space, prove that every subsequence of a Cauchy sequence is Cauchy.
7. On the real line, suppose xn → x and (x n − yn ) → 0 . Prove that yn → x .
8. Prove that the following sequences are monotone from some point. Then determine
if the sequences are bounded. If a sequence is bounded, compute its limit.
(a)
2n ∞
n + 6 n =1
{ }
 n! ∞
(b)  2 
 6n n =1
 (n − 3)2 ∞

(c) 
 n 2 
n =1
 n 7 ∞
(d)  2 n 
 e 
n =1
9. (a) Suppose xn → ∞ . Prove that {x n} is not bounded.
(b) Suppose {x n} is not bounded above. Prove that there exists a subsequence {x nk }
such that lim xnk = + ∞ .
k→ ∞
Dr. Neal, WKU
10. Let {an } and {bn} be real-valued sequences with 0 ≤ an ≤ bn for all n . Prove the
following results:
(a) If bn → 0 , then an → 0 .
(b) If an → ∞ , then bn → ∞ .
11. Let {an }, {bn} , and {c n} be real-valued sequences with an ≤ bn ≤ cn for all n .
Suppose an → x and cn → x . Prove that bn → x .
12. Use the formal definitions of limit to prove the following:
(a) lim
n →∞
9+
1
=3
n
(b) lim
n →∞
n = +∞
(c)
lim
n →∞
8
n
3/ 2
=0
13. Give the reason for convergence or divergence of each sequence.
(a)
(c)
{
{
  −2  n
(b)   
 3
∞
(−5)n
n= 1
}
sin n − cosn ∞
n!
n= 1
}
∞



n=1
 1 − 3n−2
(d) 

3n+1
∞


n =1
k
14. Let un = ( x1,n , x2,n , . . ., xk,n ) denote the n th term of a sequence in R . Prove that
k
un → u = ( x1 , x2 , . . ., x k ) in R if and only if xi,n → xi in ℜ for all i = 1, 2, . . . , k .
k
15. Apply Exercise 13 to prove that R is complete.