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Chapter 4
Sequences and Limit of
Sequences
4.1
Sequences: Basic De…nitions
De…nition 278 (sequence) A sequence is a function whose domain is a subset
of the form
fn 2 Z : n
n0 for some n0 2 Zg
The elements or the terms of the sequence, usually denoted xn will be of the
form: xn = f (n). If the terms of a sequence are denoted xn , then the sequence
is denoted (xn ) or fxn g. We can think of a sequence as a list of numbers. In this
case, a sequence will look like: fxn g = fx1 ; x2 ; x3 ; :::g. The starting point, n0 is
usually 1 but it does not have to be. However, it is understood that whatever
the starting point is, the elements xn should be de…ned for any n
n0 . For
2n
, then, we must have
example, if the general term of a sequence is xn =
n 4
n0 5.
A sequence can be given di¤erent ways.
1. List the elements. For example,
1 2 3
; ; ; ::: . From the elements listed,
2 3 4
the pattern should be clear.
n
n 2
2. Give a formula to generate the terms. For example, xn = ( 1)
. If
n!
the starting point is not speci…ed, we use the smallest value of n which
will work.
3. A sequence can be given recursively. The starting value of the sequence
is given. Then, a formula to generate the nth term from one or more
103
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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
previous terms. For example, we could de…ne a sequence by giving:
x1 = 2
1
xn+1 = (xn + 6)
2
Another example is the Fibonacci sequence de…ned by:
x1 = 1, x2 = 1
xn = xn
1
+ xn
2
for n
3
Like a function, a sequence can be plotted. However, since the domain is a
subset of Z, the plot will consist of dots instead of a continuous curve.
Since a sequence is de…ned as a function. everything we de…ned for functions
(bounds, supremum, in…mum, ...) also applies to sequences. We restate those
de…nitions for convenience.
De…nition 279 (Bounded Sequence) As sequence (xn ) is said to be bounded
above if its range is bounded above. It is bounded below if its range is bounded below. It is bounded if its range is bounded. If the domain of (xn ) is fn 2 Z : n k for some integer kg
then the above de…nition simply state that the set fxn : n kg must be bounded
above, below or both.
De…nition 280 (One-to-one Sequences) A sequence (xn ) is said to be oneto-one if whenever n 6= m then xn 6= xm .
De…nition 281 (Monotone Sequences) Let (xn ) be a sequence.
1. (xn ) is said to be increasing if xn xn+1 for every n in the domain of the
sequence. If we have xn < xn+1 , we say the sequence is strictly increasing.
2. (xn ) is said to be decreasing if xn
the sequence.
xn+1 for every n in the domain of
3. A sequence that is either increasing or decreasing is said to be monotone.
If it is either strictly increasing or strictly decreasing, we say it is strictly
monotone.
De…nition 282 (constant) (xn ) is constant if xn+1 = xn 8n. (xn ) is eventually constant if xn+1 = xn 8n > N for some positive integer N .
De…nition 283 (Periodic) (xn ) is periodic if 9p 2 Z+ such that 8n, xn+p =
xn .
Sometimes, when studying at a sequence, it is useful to look only at some
of its terms, not all of them. For example, we could take every other term. We
could also take some of the terms at random. More speci…cally, given a sequence
fan g = fa1 ; a2 ; a3 ; :::g, we can de…ne a new sequence fbk g = fank g, made from
4.2. LIMIT OF A SEQUENCE: DEFINITIONS
105
some of the terms of fan g as follows: Let n1 denote the index of the …rst element
of fan g we select and de…ne b1 = an1 .. Let n2 denote the index of the second
element of fan g we select and de…ne b2 = an2 . In general, let nk denotes the
index of the k th element of fan g we select and de…ne bk = ank . In doing so,
note that nk k (as long as n 1 which we can achieve by renumbering).
De…nition 284 (subsequence) The sequence fbk g as de…ned above is called a
subsequence of fan g. Simply speaking, a subsequence is a sequence which only
contains some of the terms of the original sequence. However, it still contains
an in…nite amount of terms, in other words, it does not stop.
Another way of de…ning a subsequence is given below.
De…nition 285 (subsequence) A subsequence of a sequence (xn ) is a sequence of the form x'(n) where ' : N ! N is a strictly increasing function.
For example, (x2n ) is a subsequence of (xn ), in this case, ' (n) = 2n. (x2n+1 )
is another subsequence, in this case ' (n) = 2n + 1.
Example 286 Given a sequence (xn )
1. (x2n ) is a subsequence of (xn ), in this case, ' (n) = 2n.
2. (x2n+1 ) is another subsequence, in this case ' (n) = 2n + 1.
4.2
Limit of a Sequence: De…nitions
4.2.1
The Concepts of Eventually and Frequently
De…nition 287 (Eventually and frequently) Let (xn ) be a sequence and
S R.
1. (xn ) is eventually in S if there exists N > 0 such that n
N =) xn 2 S.
2. (xn ) is frequently in S if xn 2 S for in…nitely many values of n.
Examples:
n
(xn ) where xn = ( 1) is frequently in the set f1g and well as frequently
in the set f 1g.
(xn ) where xn =
1
0
if 1 n 20
is eventually in the set f0g.
if
n > 20
(xn ) where xn = ln n is eventually is the set [2; 1).
The following are simple observations regarding these two notions:
If a sequence is in a set, then it is eventually in the set.
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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
If a sequence is eventually in a set, then it is frequently in the set.
(xn ) is eventually in S if and only if it is not frequently in R n S.
(xn ) is frequently in S if and only if it is not eventually in R n S.
The following is an important theorem regarding subsequences.
Theorem 288 Let (xn ) be a sequence and S
R. The following are equivalent:
1. (xn ) is frequently in S.
2. (xn ) has a subsequence in S.
Proof. We prove both directions.
1. (1 =) 2). Since (xn ) is frequently in S, (xn ) 2 S for in…nitely many n.
Let ni be the ith index of (xn ) for which (xn ) 2 S. Then the sequence (yn )
where yi = xni is both a subsequence of (xn ) and is in S.
2. (2 =) 1). Since a subsequence of (xn ) contains an in…nite number of
terms of (xn ), the result follows.
4.2.2
De…nitions of Limits
In the following, we assume that (xn ) is a sequence of real numbers and L is an
extended real number. In this section, We wish to investigate the behavior of
xn as n gets arbitrarily large, in other words as n ! 1. This is called …nding
the limit of xn as n approaches 1, it is denoted lim xn . We want to know if
n!1
the values of xn keep on changing as n ! 1, or if they seem to get closer to
some number. Several things can happen. As n ! 1, xn could get arbitrarily
large (or small), in other words, xn ! 1. Another possibility is that the
values of xn will get closer and closer to some number we will call L. But, it is
also possible that the values of xn will keep on changing forever. If we allow L
to be an extended real number, then the …rst two cases can be summarized in
lim xn = L. We now give a more precise de…nition of limits as well as partial
n!1
limits.
De…nition 289 (Limit) lim xn = L if for every neighborhood U of L, (xn )
n!1
is eventually in U . In this case, we write one of the following:
lim xn = L
n!1
lim xn = L
xn ! L as n ! 1
xn ! L
4.2. LIMIT OF A SEQUENCE: DEFINITIONS
107
If L is a real number, we say that (xn ) converges to L or that L is a …nite
limit for (xn ). Otherwise, we sat that (xn ) diverges. Note that (xn ) may diverge
for two reasons. The …rst one is if L = 1, in this case, we say that (xn ) has
an in…nite limit. The second is if (xn ) has no limit at all.
De…nition 290 (Partial limit) L is a partial limit of (xn ) if for every neighborhood U of L, (xn ) is frequently in U .
We now give equivalent conditions for a sequence to have a limit and a
partial limit. In each case, we consider whether the limit (partial limit) is …nite
or in…nite.
Theorem 291 (Finite Limit) Let (xn ) be a sequence, and L 2 R. The following are equivalent:
1. L is a limit of (xn ) in other words lim xn = L.
n!1
2. 8 > 0, (xn ) is eventually in (L
; L + ).
3. 8 > 0,9N > 0 such that n
N =) xn 2 (L
4. 8 > 0,9N > 0 such that n
N =) jxn
; L + ).
Lj < .
Proof. We prove (1 =) 2 =) 3 =) 4 =) 1).
1. (1 =) 2). Let > 0 be given. Then (L
; L + ) is a neighborhood of
L. Thus, if lim xn = L it follows by de…nition that (xn ) is eventually in
n!1
(L
; L + ).
2. (2 =) 3). This is the de…nition of eventually.
3. (3 =) 4). xn 2 (L
=) jxn Lj < .
; L + ) =) L
< xn < L + =)
< xn
L<
4. (4 =) 1). We need to show that (xn ) is eventually in U where U is any
neighborhood of L. If U is any neighborhood of L, then by de…nition, U
contains (L
; L + ) for some > 0. We can …nd N > 0 such that
n
=)
N =) jxn
< xn
=)
L
=)
xn 2 (L
=)
Lj <
L<
< xn < L +
;L + )
xn 2 U
Thus xn is eventually in U . Since U was arbitrary, the result follows.
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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Remark 292 Part 4 of theorem 291 is known as the
limit.
de…nition of the
Theorem 293 (Finite Partial Limit) Let (xn ) be a sequence, and L 2 R.
The following are equivalent:
1. L is a partial limit of (xn ).
2. 8 > 0, (xn ) is frequently in (L
; L + ).
3. 8 > 0, xn 2 (L
; L + ) for in…nitely many n.
4. 8 > 0,jxn
for in…nitely many n.
Lj <
Proof. See problems
Theorem 294 (In…nite Limit) Let (xn ) be a sequence, and L = 1 (the case
L = 1 is similar and left to the reader). The following are equivalent:
1. 1 is a limit of (xn ) in other words lim xn = 1
n!1
2. 8M > 0, (xn ) is eventually in (M; 1).
3. 8M > 0, 9N > 0 such that n
N =) xn 2 (M; 1).
4. 8M > 0, 9N > 0 such that n
N =) xn > M
Proof. See problems
Theorem 295 (In…nite Partial Limit) Let (xn ) be a sequence, and L = 1
(the case L = 1 is similar and left to the reader). The following are equivalent:
1. 1 is a partial limit of (xn ).
2. 8M > 0, (xn ) is frequently in (M; 1).
3. 8M > 0, xn 2 (M; 1) for in…nitely many n.
4. 8M > 0, xn > M for in…nitely many n.
5. (xn ) is unbounded above.
Proof. See problems
4.2. LIMIT OF A SEQUENCE: DEFINITIONS
4.2.3
109
Examples
Before we look at examples and elementary theorems, let us make some remarks.
jxn Lj simply represents the distance between xn and L. If we think of
as a very small quantity, then the above de…nition says that we can make
the distance between xn and L as small as we want, simply by taking n
large enough . In fact, no matter how small is, we can make the distance
smaller. In essence, this is saying that if n is large enough, xn and L
become identical. Think of this as a game between two persons. The
other person gives you a value for , that it the other person tells you how
close he or she wants xn to be from L. You must be able to …nd N such
that if n N , xn will be within the prescribed distance. Not only xn , but
also all the terms after xn will be within the prescribed distance.
Graphically, if we draw the horizontal lines y = L + and y = L
, then
saying n
N ) jxn Lj < means that all the values of xn will fall
within the region between the two lines whenever n N .
It should be clear to the reader that to prove the limit is L, is arbitrary.
Given any , the reader must prove the number N in the de…nition can
be found. Usually, N will depend on . We now look at some examples to
see how this is done.
Example 296 Let (xn ) be such that xn = c, for some constant c. Show that
xn ! c.
We need to show that for any > 0, (xn ) is eventually in (c
; c + ). Since
xn = c for any n, (xn ) is always in fcg (c
; c + ) for any > 0. Hence,
(xn ) is eventually in (c
; c + ).
Example 297 This example is similar to the previous one, the di¤ erence is
that (xn ) is eventually constant, that is there exists N > 0 such that xn = c
whenever n N . Again, show that xn ! c.
The proof is similar. Here, we have that (xn ) is eventually in fcg (c
; c + ).
n
Example 298 Let (xn ) be such that xn = ( 1) . Prove that (xn ) does not
have a limit (…nite or in…nite) but has two partial limits: 1 and 1.
First, we prove that 1 is not a limit of (xn ). We can …nd M > 0 such
that (xn ) is not eventually in (M; 1). Any number M > 1 will do.
Next, we prove that 1 is not a limit of (xn ).We can …nd M < 0 such
that (xn ) is not eventually in ( 1; M ). Any number M < 1 will do.
Next, we prove that no real number can be a limit of (xn ). We note that
xn only has two values, 1 and 1. For a real number x to be a limit of
(xn ), both 1 and 1 would have to be eventually in (x
; x + ), for any
1
> 0. Clearly, if = , this cannot happen, no matter what x is. So, no
2
real number can be a limit of (xn ).
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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Next, we prove that 1 is a partial limit. We need to prove that (xn ) is
frequently in (1
; 1 + ), for any > 0. By theorem 288, it is enough
to show (xn ) has a subsequence in (1
; 1 + ), for any > 0. The
subsequence (x2n ) is in f1g
(1
; 1 + ), for any > 0. The result
follows.
Finally, we prove that 1 is a partial limit. This is similar. We simply
note that the subsequence (x2n 1 ) is in f 1g.
Example 299 Let (xn ) be such that xn = n. Prove that xn ! 1. We need to
show that for any M > 0, there exists N > 0 such that n N =) xn > M . We
want xn > M . This will happen when n > M (since xn = n). So, we see that
if N is any integer greater than M , we will have n > M =) xn > M . Hence,
the result.
Example 300 Let (xn ) be such that xn = n2 . Prove that xn ! 1. We need
to show that for any M > 0, there exists N > 0 such that n N =) xn > M .
2
2
We want xp
n > M . This will happen when n > M (since xn =pn ), that is
when n > M . So, we see that if N is any integer greater than M , we will
have n > M =) xn > M . Hence, the result.
n
Example 301 Let (xn ) be such that xn = ( 1) n2 . Prove that (xn ) has no
limit and that 1 are partial limits.
1. First, we see why neither 1 nor 1 can be limits of (xn ). (xn ) cannot
be eventually in (M; 1) for any number M > 0 because if xn 2 (M; 1),
xn+1 2
= (M; 1). For the same reason, (xn ) cannot be eventually in
( 1; M ) for any number M < 0.
2. Then, we see why a real number x cannot be a limit for (xn ). Let x 2 R.
n
For x to be a limit of this sequence, ( 1) n2 has to be eventually in
(x
; x + ) for any > 0. We consider several cases.
x = 0. jxn j increases without bounds, so xn will eventually get out of
any interval of the form ( ; ).
x 6= 0. Let
there (why?)
=
jxj
. If xn is in (x
2
; x + ) then xn+1 cannot be
3. Finally, we see why 1 and 1 are partial limits of (xn ). The subsequence
(x2n ) is eventually (M; 1) for any number M > 0 thus (xn ) is frequently
in (M; 1) for any number M > 0. This means that 1 is a partial limit
of (xn ). Similarly, the subsequence (x2n+1 ) is eventually in ( 1; M ) for
any number M < 0. This means that 1 is a partial limit of (xn ).
1
Example 302 Show that the sequence whose general term is xn = converges
n
to 0.
4.2. LIMIT OF A SEQUENCE: DEFINITIONS
Given
> 0, we need to …nd N such that n
1
n
we usually start with what we want, i.e.
111
N )
1
n
0 < . To do this,
0 < , and we try to …nd the
condition n has to satisfy for it to happen.
1
n
0 < ,
,
1
<
n
,n>
So, we see that if N =
1
, then n
1
<
n
N)
1
1
n
0 < .
1
Example 303 Show that the sequence whose general term is xn = 2 converges
n
to 0.
1
Given > 0, we need to …nd N such that n N ) 2 0 < . To do this,
n
1
we usually start with what we want, i.e.
0 < , and we try to …nd the
n2
condition n has to satisfy for it to happen.
1
n2
0 < ,
,
1
<
n2
1
<
n2
, n2 >
1
1
, n > p or n <
1
p
1
1
Since n is a natural number, we only retain n > p . So, we see that if N = p ,
then n
N)
1
n
0 < .
Example 304 Same example as above, using a slightly di¤ erent approach.
1
1
1
When we had 2 < . We could have noticed that if n > 1 we have 2 < .
n
n
n
1
1
So, if we make
< then what we need, 2 < will also happen. We can
n
n
1
1
1
make < simply by taking n > , it will follow that 2 0 < . This time,
n
n
1
we …nd that N = works.
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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Remark 305 It seems that if we do the same problem di¤ erent ways, we …nd a
di¤ erent solution. We really don’t. When we …nd N , the only thing we guarantee
is that if n
N , then jxn Lj < . It might also be true if n is slightly less
than N . In other words, we are not …nding the smallest N that will work. If we
use approximations like in example 304 , we may not get the best estimate for
N.
Remark 306 The idea is the following. Forget mathematics for a second (just
a second). Imagine you are driving on a dirt road and you come to a bridge.
You are not sure if the bridge can withstand the weight of your car. If ahead of
you is a vehicle you know to be heavier than yours which just cleared the bridge,
then you know your vehicle can clear the bridge as well.
Remark 307 The above example illustrates a general technique to …nd N given
. In general, the general term of the sequence is too complicated for us to be
able to solve for n: If we call xn the general term of the sequence, we do the
following: Start with jxn Lj and simplify it as much as possible. We get an
expression involving n, call it E1 (n). If we still cannot solve for n in E1 (n) < ,
we try to …nd an upper bounds of jxn Lj, call it E2 (n) for which we can solve
E2 (n) < . The result will follow. Since jxn Lj = E1 (n) < E2 (n), if we can
have E2 (n) < , we will also have jxn Lj < . Obviously, E2 (n) has to be
small enough so we can make it as small as we want. We illustrate this with
the next example.
Example 308 Show that the sequence fxn g whose general term is xn =
2n 1
3n + 2
2
converges to .
3
Again, as before, we look at what we need, and see if we can derive the condition
n must satisfy.
2n 1
3n + 2
2
3
<
()
()
6n
3 6n 7
3 (3n + 2)
10
<
3 (3n + 2)
Though here we could solve for n, we can also use an upper bound instead. We
notice that
10
10
=
3 (3n + 2)
9n + 6
10
<
9n
10
< , it will follow that
9n
10
happens when n >
. So, given > 0, n >
9
2n 1
2
proves that lim
= .
3n + 2
3
Thus, if we make
2n 1
3n + 2
10
=)
9
2
< . The former
3
2n 1 2
< this
3n + 2 3
4.2. LIMIT OF A SEQUENCE: DEFINITIONS
113
p
p
Example 309 Prove that lim n + 1
n = 0.
Again, as before, we look at what we need, and see if we can derive the condition
n must satisfy.
p
p
p
p
p
p
n+1
n
n+1+ n
p
<
n+1
n
0 < ()
p
n+1+ n
We work on the left side of the inequality and …nd an upper bound.
p
p
p
p
n+1
n
n+1+ n
1
p
= p
p
p
n+1+ n
n+1+ n
1
p
<
2 n
p
p
1
So, if we make p < , we will have what we need, that is n + 1
n < .
2 n
p
1
1
The former happens when n >
or n > 2 (since n > 0). Now, we see
2
4
p
p
1
that given > 0, if n > 2 then
n+1
n
0 < thus proving that
4
p
p
lim n + 1
n = 0.
4.2.4
Exercises
1
. Prove that xn ! 3.
n
2
2. Let xn = 3 + . Prove that xn ! 3.
n
1
3. Let xn = for n 2 N. If x 2 R : x 6= 0, prove that x is not a partial limit
n
of (xn ).
8
n
n is a multiple of 3
< ( 1) n3 if
0
if n is one more than a multiple of 3 . Prove
4. Let xn =
:
4
if n is two more than a multiple of 3
that the partial limits of (xn ) are 1; 1; 0; 4.
1. Let xn = 3 +
3 + 2n
. Prove that xn ! 2.
5+n
8
1
>
<
if
n = 2k
2n
6. Let xn =
for any integer k. Prove that xn !
1
>
: 2
if n = 2k + 1
n +1
0.
5. Let xn =
7. Let xn =
1
n2 + 3n + 1
. Prove that xn ! .
2n2 + n + 4
2
8. Explain and justify the following observations made in the notes:
114
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
(a) If a sequence is in a set, then it is eventually in the set.
(b) If a sequence is eventually in a set, then it is frequently in the set.
(c) (xn ) is eventually in S if and only if it is not frequently in R n S.
(d) (xn ) is frequently in S if and only if it is not eventually in R n S.
9. Prove theorem 288.
10. Prove theorem 293.
11. Prove theorem 294.
12. Prove theorem 295.
4.3. LIMIT OF A SEQUENCE: THEOREMS
4.3
115
Limit of a Sequence: Theorems
These theorems fall in two categories. The …rst category deals with ways to
combine sequences. Like numbers, sequences can be added, multiplied, divided,
... Theorems from this category deal with the ways sequences can be combined
and how the limit of the result can be obtained. If a sequence can be written
as the combination of several "simpler" sequences, the idea is that it should be
easier ti …nd the limit of the "simpler" sequences. These theorems allow us to
write a limit in terms of easier limits. however, we still have limits to evaluate.
The second category of theorems deal with speci…c sequences and techniques
applied to them. Usually, computing the limit of a sequence involves using
theorems from both categories.
4.3.1
Limit Properties
We begin with a few technical theorems. They do not play an important role in
computing limits, but they play a role in proving certain results about limits.
Theorem 310 Let x be a number such that 8 > 0, jxj < , then x = 0.
Proof. See problems at the end of the section.
Theorem 311 If a sequence converges, then its limit is unique.
Proof. We do a proof by contradiction. Assume that an ! L1 and an ! L2 .
Given > 0 choose N1 such that n N1 =) jan L1 j < . Similarly, choose
2
N2 such that n
N2 =) jan L2 j < . Let N = max (N1 ; N2 ). If n
N,
2
then
jL1
L2 j = jL1
jan
<
By theorem 310 , it follows that L1
2
+
an + an
L1 j + jan
2
L2 j
L2 j
=
L2 = 0, that is L1 = L2 .
Theorem 312 If a sequence converges, then it is bounded, that is there exists
a number M > 0 such that jan j M for all n.
Proof. Choose N such that n N =) jan Lj < 1. By the triangle inequality,
we have
jan j
jLj
jan
Lj
<1
Thus, if n N , jan j < 1 + jLj. Let M1 = max (ja1 j ; ja2 j ; :::; jaN j). Let M =
max (M1 ; 1 + jLj). Then, clearly, jan j < M
116
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Theorem 313 If a sequence converges, then 8 > 09N : m; n
jan am j <
Proof. Given > 0, we can choose N such that n; m N =) jan
and jam
Lj <
jan
2
N =)
Lj <
2
. Now,
am j = jan
L+L
= j(an
L)
j(an
<
2
+
am j
(am
L)j + jam
2
L)j
Lj by the triangle inequality
=
The above theorem simply says that if a sequence converges, then the difference between any two terms gets smaller and smaller. It should also be clear
to the reader that if an ! L, then so does an+k where k is any natural number.
The above theorem can be used to prove that a sequence does not converge
by proving that the di¤erence between two of its terms does not get smaller and
smaller.
Example 314 Find lim cos n
We suspect the sequence diverges, as its values will oscillate between -1 and
1 . We can actually prove it using theorem 313. We notice that for any n,
jcos 2n
cos ((2n + 1) )j = 2. For the sequence to converge, this di¤ erence
should approach 0. Hence, the sequence diverges.
Theorem 315 Suppose that (an ) converges. Then, any subsequence (ank ) also
converges and has the same limit.
Proof. Suppose that an ! L. Let bk = ank be a subsequence. We need to prove
that given > 0, there exists N such that k
N =) jbk Lj < . Let > 0
be given. Choose N such that nk N =) jank Lj < . Now, if k N , then
nk N therefore
jbk
Lj = jank
Lj
<
This theorem is often used to show that a given sequence diverges. To do so,
it is enough to …nd two subsequences which do not converge to the same limit.
Alternatively, once can …nd a subsequence which diverges.
Example 316 Study the convergence of cos n
The subsequence cos 2n converges to 1, while the subsequence cos (2n + 1)
converges to 1. Thus, cos 2n must diverge.
In the next two sections, we look at theorems which give us more tools to
compute limits.
4.3. LIMIT OF A SEQUENCE: THEOREMS
4.3.2
117
Limit Laws
The theorems below are useful when …nding the limit of a sequence. Finding the
limit using the de…nition is a long process which we will try to avoid whenever
possible. Since all limits are taken as n ! 1, in the theorems below, we will
write lim an for lim an .
n!1
Theorem 317 Let (an ) and (bn ) be two sequences such that an ! a and bn ! b
with a and b real numbers. Then, the following results hold:
1. lim (an
bn ) = (lim an )
(lim bn ) = a
b
2. lim (an bn ) = (lim an ) (lim bn ) = ab
an
bn
3. if lim bn = b 6= 0 then lim
=
lim an
a
=
lim bn
b
4. lim jan j = jlim an j = jaj
5. if an
0 then lim an
6. if an
bn then lim an
7. if lim an = a
0
lim bn
p
p
p
0 then lim an = lim an = a
Proof. We prove some of these items. The remaining ones will be assigned as
problems at the end of the section.
1. We prove
(lim an )
9N : n
such that
jbn
lim (an + bn ) = (lim an ) + (lim bn ). The proof of lim (an bn ) =
(lim bn ) is left as an exercise. We need to prove that 8 > 0,
N =) jan + bn (a + b)j < . Let > 0 be given, choose N1
n
N1 =) jan aj < . Choose N2 such that n
N2 =)
2
bj < . Let N = max (N1 ; N2 ). If n N , then
2
jan + bn
(a + b)j = j(an
jan
<
2
+
a) + (bn
aj + jbn
2
b)j
bj by the triangle inequality
=
2. We need to prove that 8 > 0, 9N : n
N =) jan bn abj < . Since
an converges, it is bounded, let M be the bound i.e. jan j < M . Choose
N1 such that n
N1 =) jan aj <
. Choose N2 such that
2 (jbj + 1)
118
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
n
N2 =) jbn
bj <
jan bn
2 (M + 1)
. Let N = max (N1 ; N2 ). If n
abj = jan bn
an b + an b
= jan (bn
<M
<
2
bj + jbj jan
2 (M + 1)
+
2
abj
b) + b (an
jan j jbn
+ jbj
N then
a)j
aj
2 (jbj + 1)
=
3. See problems
4. We need to prove that jan j ! jaj that is 8 > 0, 9N : n
jjan j jajj < . Let > 0 be given, choose N such that n
jan aj < (since an ! a) . If n N , then we have:
jjan j
jajj < jan
N =)
N =)
aj by the triangle inequality
<
5. We prove it by contradiction. Assume that an ! a < 0. Choose N such
1
a. Then,
that n N ) jan aj <
2
an
1
a
2
a<
1
a
2
) an < 0
) an <
which is a contradiction.
6. We apply the results found in parts 1 and 5 to the sequence an
bn .
7. See problems
Remark 318 Parts 1, 2 and 3 of the above theorem hold even when a and b
are extended real numbers as long as the right hand side in each part is de…ned.
You will recall the following rules when working with extended real numbers:
1. 1 + 1 = 1
2.
1
1 = ( 1) ( 1) = 1
1 = ( 1) 1 = 1 ( 1) =
3. If x is any real number, then
(a) 1 + x = x + 1 = 1
1
4.3. LIMIT OF A SEQUENCE: THEOREMS
119
(b)
1+x=x 1= 1
x
x
(c)
=
=0
1
1
x
1 if x > 0
(d)
=
1 if x < 0
0
(e) 1
x=x
(f ) ( 1)
1 if x > 0
1 if x < 0
1=
x=x
1 if
1 if
( 1) =
x>0
x<0
4. However, the following are still indeterminate forms. Their behavior is
unpredictable. Finding what they are equal to requires more advanced techniques such as l’Hôpital’s rule.
(a)
1 + 1 and 1
(b) 0 1 and 1
0
1
and
(c)
1
0
0
1
Remark 319 When using theorems from this category, it is important to remember previous results since these theorems allow us to write a limit in terms
of other limits, we hopefully know. The more limits we know, the better o¤ we
are.
Example 320 If c 6= 0, …nd lim nc . We know from an example in the previous
section that lim n1 = 0. Therefore
lim
c
n
= c lim
= c
=
1
n
0
0
Example 321 Find lim n12 . In the previous section, we computed this limit
using the de…nition. We can also do it as follows.
lim
1
n2
1
n
1
=
lim
n
= 0 0
=
=
lim
1
n
lim
1
n
0
Remark 322 From the above example, we can see that if p is a natural number,
lim n1p = 0.
120
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
2
Example 323 Find lim n2n+3n
2 +1 .
This problem involves using a standard technique you should remember. We
show all the steps, then we will draw a general conclusion. We begin by factoring
the term of highest degree from both the numerator and denominator.
lim
n2 + 3n
2n2 + 1
=
lim
n2 1 + n3
n2 2 + n12
=
lim
1 + n3
2 + n12
=
lim (1) + 3 lim
=
1
Now,
lim 1 +
3
n
1
n
and
lim 2 +
1
n2
=
lim (2) + lim
=
2
1
n2
Since both the limit of the numerator and denominator exist and the limit of the
denominator is not 0, we can write
lim
n2 + 3n
2n2 + 1
=
=
lim 1 + n3
lim 2 + n12
1
2
Remark 324 The same technique can be applied to every fraction for which
the numerator and denominator are polynomials in n. We see that the limit
of such a fraction will be the same as the limit of the quotient of the terms of
highest degree.
4.3.3
More Theorems on Limits
In example 304 , we used an approximation to simplify the problem a little bit.
In this particular example, the approximation was not really necessary, it was
more to illustrate a point. Sometimes, if the problem is more complicated, it
may be necessary to use such an approximation in order to be able to …nd the
condition n has to satisfy. In other words, when we try to satisfy jxn Lj < ,
we usually simplify jxn Lj to some expression involving n. Let E1 (n) denote
this expression. This gives us the inequality E1 (n) < which we have to solve
for n. If it is too hard, we then try to …nd a second expression we will call E2 (n)
such that E1 (n) < E2 (n) < . E2 (n) should be such that solving the inequality
E2 (n) < is feasible and easier. In order to achieve this, several tricks are used.
We recall some useful results, as well as some theorems below.
4.3. LIMIT OF A SEQUENCE: THEOREMS
121
Theorem 325 (Bernoulli’s inequality) If x
n
ber, then (1 + x)
1 + nx.
1, and n is a natural num-
n
Theorem 326 (binomial theorem) (a + b) = an +nan
n (n
1) (n
3!
2)
an
3 3
b + ::: + nabn
1
n
Corollary 327 (1 + x) = 1 + nx +
nxn
1
+ xn .
1
b+
n (n 1) n
a
2
2 2
b +
+ bn .
n (n 1) 2 n (n
x +
2
1) (n
3!
2)
x3 + ::: +
n
In particular, when x 0, then (1 + x) is greater than any part of the right
n
hand side. For example, we obtain Bernoulli’s inequality: (1 + x)
1 + nx.
n
(n
1)
(n
2) 3
n
(n
1)
n
n
x2 or (1 + x)
x .
We could also write (1 + x)
2
3!
n
And so on. This is useful to get approximations on quantities like 3 . We
rewrite it as
n
3n = (1 + 2)
Theorem 328 (squeeze theorem) If an ! L, cn ! L and an
bn
cn ,
then bn ! L
Proof. We need to prove that 8 > 09N : n N =) jbn Lj < . Let > 0
be given. Choose N1 such that n N1 =) jan Lj < or
< an L < .
Similarly, choose N2 such that n
N2 =)
< cn L < . Let N =
max (N1 ; N2 ). If n N then
an
bn
()
()
() jbn
cn () an
< an
L
< bn
L<
L
bn
L
cn
bn
L
cn
L<
L
Lj <
Theorem 329 If 0 < a < 1 then an ! 0
1
1
1. Then, x > 0 and a =
. For n
Proof. Let x =
a
1+x
1
an =
n
(1 + x)
1
by Bernoulli’s inequality
1 + nx
1
<
nx
To show that an ! 0, we need to show that jan j < , for any
that is
1
an < ()
<
nx
1
() n >
x
1,
whenever n
N,
122
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
So, given
> 0, N; =
1
will work.
x
We look at a few more examples, to see how all these results come into play.
2n2 + 1
using the de…nition
n2 + 1
Of course, we can …nd this limit by using the theorems on limits. Here, we do
it using the de…nition, as asked. We think the limit is 2. We want to show that
2n2 + 1
for every > 0, there exists N such that n N =)
2 < . First,
n2 + 1
we simplify the absolute value.
Example 330 Find lim
2n2 + 1
n2 + 1
1
+1
1
= 2
n +1
1
< 2
n
1
< if n > 1
n
2 =
n2
1
1
< , which happens when n > , then we will have
n
1
2 < . So, N = will work.
So, we see that if
2n2 + 1
n2 + 1
1
n2 + 2n 4
We think the limit is 0. We need to prove that for every > 0, there exists N
1
such that n N =) 2
< . We begin by noticing that
n + 2n 4
Example 331 Find lim
n2 + 2n
4 = n2 + 2 (n
2)
2
> n if n > 2
Therefore, if n > 2,
n2
1
+ 2n
4
=
n2
1
+ 2n
4
1
< 2
n
1
<
n
If n > max 2;
1
, then
n2
1
+ 2n
4
< . So, N = max 2;
1
will work.
Remark 332 The two examples above could have been done using the techniques discussed in the previous subsection, that is without using the de…nition.
4.3. LIMIT OF A SEQUENCE: THEOREMS
123
4n
Example 333 Prove that lim
=0
n!
The trick we use is worth remembering. We will use the squeeze theorem. We
note that
4 4 4 ::: 4
4n
=
n!
1 2 3 ::: n
43
4
4 4 ::: 4
since
3!
n
4 5 ::: n 1
0
3
But lim 43!
4
n
1
= 0 hence the result follows by the squeeze theorem.
4n
2n
4n
We suspect this sequence diverges to in…nity. We need to prove that
grows
2n
without bounds, that is for every M > 0, there exists N such that n N =)
4n
> M . First, we notice that
2n
Example 334 Find lim
n
4n
(1 + 3)
=
2n
2n
n (n 1) 32
2
>
by using the binomial theorem
2n
9 (n 1)
>
4
We can see that
9 (n 1)
4M
> M () n 1 >
4
9
4M + 9
() n >
9
so, N =
4M + 9
will work.
9
n (n 1) 32
n
Remark 335 We used the binomial theorem to deduce that (1 + 3)
.
2
Often, when using this theorem, students do not know which term to keep, the
term in x, or x2 , or x3 ,... The answer is that it depends on what we are trying
4n
> M for any M , in other words,
to achieve. Here, we wanted to show that
2n
n
4
can be made as big as we want simply by taking n large enough. Since we
2n
4n
could not solve for n, we replaced
by a smaller term. That smaller term
2n
should have similar properties, in other words, the smaller term should also get
arbitrarily large. If we take the smaller term too small, it will not work. For
124
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
4n
1
1
>
. However,
cannot be made as large as
2n
2n
2n
n (n 1) 32
n
we want, so this does not help us. If instead of using (1 + 3)
2
n
(terms of second degree), we had used (1 + 3)
1 + 3n, then we would have
obtained
example, it is true that
1 + 3n
2n
>
()
()
()
M () 2nM < 1 + 3n
2nM
3n < 1
n (2M
3) < 1
1
n<
2M 3
So, this does not work. This tells us that 1+3n
2n > M only for a few values of n.
This is why we used terms of higher degree.
p
p
Example 336 Find lim n + 2
n
p
p
p
p
p
p
n+2
n
n+2+ n
p
lim n + 2
n = lim
p
n+2+ n
2
= lim p
p
n+2+ n
2
p
=
p
lim n + 2 + n
=0
Remark 337 In many proofs or problems, di¤ erent versions of the triangle
inequality are often used. As a reminder, here are the di¤ erent versions of the
triangle inequality students should remember.
jjaj
jbjj
ja
bj
jaj + jbj
jjaj
jbjj
ja + bj
jaj + jbj
and
Finally, we give a theorem which generalizes some of the examples we did
above.
Theorem 338 All limits are taken as n ! 1.
1
= 0.
np
p
2. If p > 0 then lim n p = 1.
p
3. lim n n = 1.
1. If p > 0 then lim
4. If p > 1 and
2 R then lim
n
= 0.
pn
4.3. LIMIT OF A SEQUENCE: THEOREMS
125
5. If jpj < 1 then lim pn = 0
6. 8p 2 R, lim
pn
= 0.
n!
Proof. We prove each part separately.
1. We can use the de…nition. Let > 0 be given. We show there exists
1
N > 0 : n N =) p 0 < . We begin with what we want to achieve.
n
1
np
<
0
1
<
np
1
()
()
np >
()
1
n> p
p
So, we see that if N is the smallest ineteger larger than
follow.
1
p
p
, the result will
2. See problems
p
3. Let xn = n n
p 1. Proving the result amounts to proving that lim xn = 0.
From xn = n n 1, we can write
p
n
n = xn + 1
n
=
(xn + 1)
n
Using the binomial theorem, we see that if n
n
n (n 1) 2
xn
2
Thus
2
x2n
It follows that for all n
Since lim
2
n 1
n
1
2 we have
0
q
2
r
xn
2
n
1
= 0, it follows using the squeeze theorem that lim xn = 0.
4. Here again, we will use the binomial theorem. Since p > 1, we can write
p = 1 + q with q > 0. Therefore, if k is a positive integer such that k > ,
we have
pn
=
n
(1 + q)
n (n 1) ::: (n
>
k!
k + 1)
qk
126
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Now, if n > 2k, then k < 12 n. It follows that n k + 1 > 12 n + 1 > 21 n. It
follows that
n (n 1) ::: (n k + 1)
nk
> k
k!
2 k!
and therefore
2k k!
1
n
0
pn
qk
nk
Since
2k k!
n!1 q k
lim
1
nk
=
=
2k k!
1
lim k
k
q n!1 n
0
It follows from the squeeze theorem that lim
5. See problems (hint: write p =
= 0).
1
q
n
=0
pn
and use part 4 of the theorem with
6. See problems.
4.3.4
Exercises
1. Prove that
5n
! 0.
n!
2. Prove that
n!
! 0.
nn
3. Finish proving theorem 338.
4. Consider (xn ) a sequence of real numbers such that lim xn = x where
n!1
x > 0, prove there exists an integer N > 0 such that n N =) xn > 0.
5. Consider (xn ) a sequence of real numbers such that xn 0 for any n. If
x is a partial limit of (xn ), prove that x
0. Prove the same result if
x = lim xn . Use this to show that if xn yn then lim xn lim yn .
6. Consider (xn ) a sequence of real numbers and let x 2 R. Prove the two
conditions below are equivalent.
(a) xn ! x as n ! 1.
(b) jxn
xj ! 0 as n ! 1.
7. Consider (xn ) a sequence of real numbers and let x 2 R. If lim xn = x,
prove that lim jxn j = jxj.
4.3. LIMIT OF A SEQUENCE: THEOREMS
127
8. Consider (xn ) a sequence of real numbers and let x 2 R. Suppose that
lim xn = x. De…ne yn = xn+p for some integer p. Prove that lim yn = x.
9. Given that an
1.
bn for every n and that lim an = 1, prove that lim bn =
10. Suppose that (an ) and (bn ) are sequences of real numbers such that
jan bn j < 1 for every n. Prove that if 1 is a partial limit of (an )
then 1 is also a partial limit of (bn ).
11. Two sequences are said to be eventually close if 8 > 0, 9N > 0 : n
N =) jan bn j < .
(a) Prove that if two sequences (an ) and (bn ) are eventually close and if
a number x is the limit of the sequence (an ) then x is also the limit
of the sequence (bn ).
(b) Prove that if two sequences (an ) and (bn ) are eventually close and if
a number x is a partial limit of the sequence (an ) then x is also a
partial limit of the sequence (bn ).
12. Determine if the sequences below have limits. In each case, if a limit L
exists, given > 0, …nd N such that n N =) jxn Lj < .
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
n 1
n+1
3n2 + 1
n2 + 1
5n3 + n 4
2n3 + 3
n3
n+1
4n2 + 1
n3 + n
1
2
n
+ 2 + ::: + 2
n2
n
n
1
2n 1
( 1)
1
2n
1
1
2
1
1
3
1
1
4
::: 1
1
n
5n
2n
p
(j) Prove that if a > 1 then a2 > a > a > 1
(i)
(k) Prove that
n
n2
n3
n4
!
0,
!
0,
!
0,
!0
2n
2n
2n
2n
128
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
13. Prove theorem 310
14. Finish proving theorem 317
15. Let (xn ) and (yn ) be two sequences. Assume further that lim xn = x and
y is a partial limit of (yn ). Prove that x + y is a partial limit for (xn + yn ).
16. State and prove similar results for subtraction, multiplication and division.
17. Give an example of two divergent sequences (xn ) and (yn ) such that
(xn + yn ) is convergent.
18. Give an example of two sequences (xn ) and (yn ) such that xn ! 0, yn !
1 and:
(a) xn yn ! 0
(b) xn yn ! c where 0 < c < 1
(c) xn yn ! 1
(d) (xn yn ) is bounded but has no limit.
19. Let (xn ) be a sequence of real numbers such that xn ! 0. Prove that
x1 + x2 + ::: + xn
7! 0
n
20. Let (xn ) be a sequence of real numbers such that xn ! x for some real
number x. Prove that
x1 + x2 + ::: + xn
7! x
n
21. Show by examples that if jxn j ! jxj then xn does not necessarily converge.
If it does converge, it does not necessarily converge to x.
22. Prove that jxn j ! 0 () xn ! 0.
23. Show by examples that if (xn ) and (yn ) are divergent sequences then
(xn + yn ) is not necessarily divergent. Do the same for (xn yn )
24. Show that if (xn ) ! 0 and (yn ) is bounded then (xn yn ) ! 0.
25. Prove that if (xn ) is a sequence in a set S
limit of (xn ) must belong to S.
26. Prove that if S
that xn ! x.
27. Let S
R, then every …nite partial
R and x 2 S then there exists a sequence (xn ) in S such
R. Prove that the following conditions are equivalent:
(a) S is unbounded above.
(b) There exists a sequence (xn ) in S such that xn ! 1.
4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES
4.4
129
Monotone Sequences and Cauchy Sequences
4.4.1
Monotone Sequences
The techniques we have studied so far require we know the limit of a sequence
in order to prove the sequence converges. However, it is not always possible
to …nd the limit of a sequence by using the de…nition, or the limit rules. This
happens when the formula de…ning the sequence is too complex to work with.
It also happens with sequences de…ned recursively. Furthermore, it is often the
case that it is more important to know if a sequence converges than what it
converges to. In this section, we look at two ways to prove a sequence converges
without knowing its limit.
We begin by looking at sequences which are monotone and bounded. These
terms were de…ned at the beginning of this chapter.
You will recall that in order to show that a sequence is increasing, several
methods can be used.
1. Direct approach, simply show that an+1
2. Equivalently, show that an+1
3. Equivalently, show that
positive.
an
an+1
an
an for every n.
0 for every n.
1 for every n if both an and an+1 are
4. If an = f (n), one can show a sequence (an ) is increasing by showing that
f is increasing that is by showing that f 0 (x) 0.
5. By induction.
We now state and prove an important theorem about the convergence of
increasing sequences.
Theorem 339 An increasing sequence (an ) which is bounded above converges.
Furthermore, lim an = sup fan g.
n!1
Proof. We need to show that given > 0, there exists N such that n N =)
jan Aj < where A is the limit. Let > 0 be given. Since (an ) is bounded
above, by theorem 158, fan g has a supremum. Let A = sup fan g. Let > 0 be
given. then, A
< A. By theorem 151, there exists an element of fan g, call
it aN , such that A
< aN
A. Since (an ) is increasing, we have an
aN
for every n N . Therefore, if n N ,
A
< an
() 0
=) jan
A ()
A
an <
Aj <
< an
A
0
130
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
Corollary 340 A decreasing sequence (an ) which is bounded below converges.
Furthermore, lim an = inf fan g.
n!1
Proof. The proof is similar to the proof of the theorem.
Theorem 341 A monotone sequence converges if and only if it is bounded.
Proof. The proof follows from results proven in the previous section and the
previous theorem and its corollary. Suppose we have a monotone sequence (xn ).
If we assume that (xn ) converges, then it follows that it is bounded by theorem
312 that it is bounded. Conversely, if we assume that (xn ) is bounded then
it is both bounded above and below. Since (xn ) is monotone, then it is either
increasing or decreasing. If it is increasing, it will converge by the previous
theorem since it is also bounded above. If it is decreasing, then it will converge
by the corollary since it is bounded below.
Example 342 Prove that the sequence whose general term is an =
n 1
P
conk=0 k!
verges.
We try to establish this result by showing that this sequence is non-decreasing
and bounded above.
(an ) is increasing: an = 1 + 1 +
an+1
1
1
1
+ + ::: + . Therefore,
2 3!
n!
an =
1
>0
(n + 1)!
(an ) is bounded above. For this, we use the fact that n! 2n
n 1. The proof of this fact is left as an exercise. Therefore,
1+1+
1
1
1
1
1
1
+ + ::: +
< 1 + 1 + + 2 + ::: + n
2 3!
n!
2 2
2
0
1
1
1
1
<1+
+
+
2
2
2
n
1
1
2
<1+
1
1
2
1
<1+
1
1
2
<3
1
for every
1
2
+ ::: +
1
2
n 1
Since (an ) is bounded above and increasing, it must converge. We will call
the limit e0 .
4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES
131
Example 343 Find lim an where (an ) is de…ned by:
n!1
a1 = 2
1
an+1 = (an + 6)
2
If we knew the limit existed, …nding it would be easy. We must …rst establish that
it exists. We do this by showing that this sequence is increasing and bounded
above. This part is left as an exercise. Once this fact has been established, then
we know the sequence must converge. Let L be its limit, we must …nd L. Before
proceeding, we will recall the following fact: lim an+1 = lim an . Therefore, if for
1
every n we have an+1 = (an + 6), then we must also have:
2
lim an+1 = lim
1
(an + 6)
2
1
(lim an + 6)
2
1
L = (L + 6)
2
2L = L + 6
L=
L=6
n
1
. Show lim an exists. This limit is in fact
n
the number e, but we won’t show that. Again, to show that (an ) converges, we
show that it is increasing and bounded above.
Example 344 Let an =
1+
(an ) bounded above. To establish this, we use the following fact: If k is
an integer such that 1 < k n, then
n (n
1) (n
2) ::: (n
nk
k + 1)
=
1
1
n
1
2
n
::: 1
k
1
n
The proof of this is left as an exercise. Using the binomial theorem, we
have:
2
1
n (n 1) 1
n (n
+
+ ::: +
n
2!
n
1
1
1
1
=1+1+
1
+ ::: +
1
2!
n
k!
n
1
1
< 1 + 1 + + ::: +
2!
n!
< e0
an = 1 + n
1) (n
1
2) ::: (n
k!
2
n
::: 1
k + 1)
k
1
n
1
n
k
+ ::: +
+ ::: +
where e0 is the limit found in the exercise above. Thus, (an ) is bounded by
e0 .
1
n!
1
n!
n!
1
n
1
n
n
1
2
n
::: 1
n
n
132
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
(an ) increasing. The kth term in the expansion of an is
1
k!
1
1
n
1
2
n
1
2
1
1
1
::: 1
k!
n+1
n+1
j
j
>1
. Hence, an+1 > an .
n+1
n
The kth term in the expansion of an+1 is
j
j
< , it follows that 1
n+1
n
So, (an ) is increasing.
Since
Since (an ) is increasing and bounded above, it follows that (an ) converges.
We de…ne this limit to be the number e.
4.4.2
Cauchy Sequences
De…nition 345 (Cauchy Sequence) A sequence (xn ) is said to be a Cauchy
sequence if for each > 0 there exists a positive integer N such that m; n
N =) jxm xn j < .
We begin with some remarks.
Remark 346 These series are named after the French mathematician Augustin
Louis Cauchy (1789-1857).
Remark 347 It is important to note that the inequality jxm xn j < must be
valid for all integers m; n that satisfy m; n N . In particular, a sequence (xn )
satisfying jxn+1 xn j < for all n N may not be a Cauchy sequence.
Remark 348 A Cauchy sequence is a sequence for which the terms are eventually close to each other.
Remark 349 In theorem 313, we proved that if a sequence converged then it
had to be a Cauchy sequence. In fact, as the next theorem will show, there is a
stronger result for sequences of real numbers.
We now look at some examples.
1
Example 350 Consider (xn ) where xn = . Prove that this is a Cauchy
n
sequence.
Let > 0 be given. We want to show that there exists an integer N > 0 such
that m; n < N =) jxm xn j < . That it we would like to have
jxn
Since
xm j < ()
1
n
1
n
1
<
m
1
1
1
< +
m
n m
1
1
1
+
< , the result will follow. This will happen if both
<
n m
n
2
2
1
2
that is when n > and
< that is when n > . So, we see that if N is an
m
2
2
integer larger than then m; n > N =) jxm xn j < .
If we make
::: 1
k
k 1
.
n+1
1
n
.
4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES
133
n
X
1
Example 351 Consider (xn ) where xn =
k2
k=1
Let > 0 be given. We want to show that there exists an integer N > 0 such
that m; n < N =) jxm xn j < (without loss of generality, let us assume that
n > m). That it we would like to have
jxn
xm j
<
()
n
X
()
k=m+1
n
X
1
k2
k=1
xm j =
=
<
n
X
k=m+1
n
X
k=m+1
n
X
k=m+1
=
<
k=1
1
<
k2
Remembering telescoping sums and the fact that
jxn
m
X
1
<
k2
1
1
<
, we see that
k2
k (k 1)
1
k2
1
k2
1
k
1
1
k
1
1
m n
1
1
+
m n
So, as we saw in the previous example, if N is an integer larger than
m; n < N =) jxm
xn j < .
2
, then
We now look at important properties of Cauchy sequences.
Theorem 352 Every Cauchy sequence is bounded.
Proof. See problems.
Theorem 353 A sequence of real numbers converges if and only if it is a
Cauchy sequence.
Proof. We have already proven one direction. Let (xn ) be a sequence of real
numbers.
If (xn ) converges, then we know it is a Cauchy sequence by theorem 313.
Assume (xn ) is a Cauchy sequence. We must prove that it converges. By
theorem 352, we know that fxn g is bounded. Therefore, by the completeness axiom, for each n, the number
an = inf fxn ; xn+1 ; xn+2 ; :::g
134
CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES
exists. Furthermore, the sequence (an ) is increasing and bounded (why?).
Therefore, by theorem 341, (an ) converges. Let a = lim an . We prove
that lim xn = a. Let > 0 be given. Since (xn ) is a Cauchy sequence, we
can …nd N > 0 such that m; n
N =) jxn xm j < : It follows that
2
i
h
(why).
both (xn ) and (an ) are contained in the interval xN
; xN +
2
2
Thus, the limit a is also in this interval. Therefore, for all n
N , we
have:
jxn
aj
<
jxn
xN j + jxN
aj
This proves that lim xn = a, thus (xn ) converges.
Remark 354 The key to this theorem is that we are dealing with a sequence of
real numbers. The fact that a Cauchy sequence of real number converges is linked
to the fact that R is complete. In fact, this is sometimes used as a de…nition of
completeness. Some texts say that a set is complete if every Cauchy sequence
converges in that set. It is possible to …nd a Cauchy sequence of rational numbers
which does not converge in Q.
We …nish this section with an important theorem.
De…nition 355 A nested sequence of intervals is a sequence fIn g of intervals with the property that In+1 In for all n.
Theorem 356 (Nested Intervals) If f[an ; bn ]g is a nested sequence of closed
intervals then there exists a point z that belongs to all the intervals. Furthermore,
if lim an = lim bn then the point z is unique.
Proof. We provide a sketch of the proof and leave the details as homework.
First, prove that fan g and fbn g are monotone and bounded. Thus, they must
converge. Let a = lim an and b = lim bn . Then, explain why a b. Conclude
from this.
4.4.3
Exercises
1. Prove that n!
2n
1
for every n
1
2. Prove that if k is an integer such that 1 < k
n (n
1) (n
2) ::: (n
nk
k + 1)
=
1
1
n
n, then
1
2
n
::: 1
k
1
n
3. Prove that the sequence given by
a1 = 2
1
an+1 = (an + 6)
2
is increasing and bounded above by 6. (hint: use induction for both).
4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES
135
4. Show that the sequence de…ned by
a1 = 1
an+1 = 3
1
an
is increasing and satis…es an < 3 for all n. Then, …nd its limit.
5. Let a and b be two positive numbers such that a > b. Let a1 be their
a+b
arithmetic mean, that is a1 =
. Let b1 be their geometric mean, that
2
p
p
an + bn
is b1 = ab. De…ne an+1 =
and bn+1 = an bn .
2
(a) Use mathematical induction to show that an > an+1 > bn+1 > bn .
(b) Deduce that both (an ) and (bn ) converge.
(c) Show that lim an = lim bn . Gauss called the common value of these
limits the arithmetic-geometric mean.
6. Prove theorem 352.
7. Answer the why? parts in the proof of theorem 353.
8. Finish proving theorem 356.