Download 2 Sequences of real numbers

2
Sequences of real numbers
Outline:
• The limit of a sequence of real numbers, bounded/monotone/Cauchy/convergent sequences
• Cauchy’s theorem for sequences
• Passing to the limit in inequalities
• Sufficient conditions for convergence: convergence of monotone sequences, squeeze theorem
• Operations with sequences which have a limit
• Césaro-Stolz theorem and the Root Criterion for sequences
Definition 2.1 A sequence of real numbers is a function  : N → R.
For the convenience of the notation, it is customary to identify the function (sequence)  with its values 0 =
 (0)  1 =  (1)  2 =  (2)  3 =  (3)  . We will adopt this convention, and we will henceforth write sequences
as ( )∈N , ( )  or simply  .
An important notion in the study of sequences is that of convergence. Intuitively, a sequence of numbers
converge to a certain number if the terms of the sequence “crowd” around that number. For example, the terms
of the sequence  = 1 are 1 12  13  14     and intuitively you can see that these numbers crowd around the value 0
(they become closer and closer to 0). The precise definition is the following.
Definition 2.2 (convergent sequence, limit of a sequence) We say that the sequence ( )∈N converges to
 ∈ R if
for any   0 there exists  () ∈ N such that | − |   for any  ≥  () 
The number  is called the limit of the sequence ( )∈N , and we write lim→∞  =  or  −→ .
→∞
Remark 2.3 If for any   0 there exists  () ∈ N such that  ≥  for any  ≥  (), we say that the sequence
 has the limit +∞. Similarly, if for any   0 there exists  () ∈ N such that  ≤ − for any  ≥  (), we
say that the sequence  has the limit −∞. We write lim→∞  = +∞ respectively lim→∞  = −∞, but we do
not say that the sequence  is convergent (this requires the limit to be finite! We may say that the sequence 
diverges to +∞, respectively −∞).
As we will see below, the notion of convergent sequence is closely related to that of Cauchy sequence, as defined
below.
Definition 2.4 (Cauchy sequence) We say that ( )∈N is a Cauchy sequence if
for any   0 there exists  () ∈ N such that | −  |   for any   ≥  () 
The number  is called the limit of the sequence ( )∈N , and we write lim→∞  =  or  −→ .
→∞
Some of the terminology used for sequences is contained in the following.
Definition 2.5 Given a sequence ( )∈N , we say that the sequence is:
a) bounded above, if there exists  ∈ R such that  ≤  , for any  ∈ N;
b) bounded below, if there exists  ∈ R such that  ≤  , for any  ∈ N;
c) bounded, if it is bounded above and below;
d) increasing, if  ≤ +1 for any  ∈ N (if the previous inequality is strict, we say that the sequence is strictly
increasing);
e) decreasing, if  ≥ +1 for any  ∈ N (if the previous inequality is strict, we say that the sequence is strictly
decreasing);
f ) monotone, if it is either increasing or decreasing.
With this preparation, we can now show that if a sequence has a limit, then the limit is unique (it is not possible
for a sequence to have two different limits).
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Proposition 2.6 If the limit of a sequence exists, it is unique.
Proof. Assume there exists 1  2 ∈ R such that the sequence ( )∈N converges to 1 and 2 . Given   0, there
exists 1 () and 2 () such that
| − |  
 ≥ 1 () 
and
| − |  
 ≥ 2 () 
It follows that we have
|1 − 2 | =
≤

=
|1 −  − (2 −  )|
| − 1 | + | − 2 |
+
2
for any  ≥ max {1 ()  2 ()} 
Thus |1 − 2 |  2 for any   0, and therefore |1 − 2 | = 0, that is 1 = 2 .
A subsequence of a given sequence ( )∈N is a sequence ( )∈N , where  are natural numbers with 0 ≤ 1 
2   For example a subsequence of the sequence  =  (that is 0 1 2 3 4 5   ) is 2 = 2 (that is the even
terms of the original sequence, 0 2 4 6   ); note that there are many possible subsequences of a given sequence:
2+1 = 2 + 1 and 2 = 2 are for example two other subsequences of the sequence  =  (the subsequence of
odd terms, respectively the subsequence of perfect squares of the sequence).
A first connection between Cauchy and convergent sequences is contained in the following.
Proposition 2.7 Let ( )∈N be a sequence of real numbers. We have the following:
a) If ( )∈N is a convergent sequence, then ( )∈N is also a Cauchy sequence.
b) If ( )∈N converges to , then any subsequence ( )∈N converges to .
c) If ( )∈N is a Cauchy sequence which has a convergent subsequence, then ( )∈N is also a convergent
sequence.
Lemma 2.8 (Cesaro’s Lemma) Any bounded sequence contains a convergent subsequence.
From the previous proposition and lemma, we obtain the following:
Theorem 2.9 (Cauchy’s theorem for sequences) A sequence is convergent if and only if it is Cauchy.
Remark 2.10 We can use the previous theorem to show for example that the sequence ( )∈N with  = 1 + 12 +
 + 1 is not convergent. By contradiction, if ( )∈N were convergent, then it were also Cauchy, and therefore
| −  | can be made arbitrarily small for all  and  sufficiently large. But
¯
¯
¯ 1
1
1 ¯¯
¯
|2 −  | = ¯
+
+  +
+1 +2
2 ¯
1
1
1
+
+  +
=
+1 +2
2
1
1
1

+
+  +
2 2
2
1
=

2
for any  ≥ 1, which shows that ( )∈N is not Cauchy, and therefore not convergent.
The following proposition gives sufficient conditions for the convergence of a sequence:
Proposition 2.11 (Convergence of monotone and bounded sequences) If the sequence ( )∈N is increasing and bounded above (or decreasing and bounded below), then it is convergent.
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Proof. Suppose that the sequence ( )∈N is increasing, and there exists  ∈ R such that  ≤  for any  ∈ N.
Let  = sup∈N  ∈ R (note that since  ≤  , we have  ≤   +∞).
By the definition of the supremum, given   0, there exists  () ∈ N such that
 −    () 
Since the sequence  is increasing, for any  ≥  () we have
 −   () ≤  ≤    + 
or equivalent
| − |  
 ≥  () 
which shows that the sequence ( )∈N is convergent to 
Similar proof for the case when the sequence ( )∈N is decreasing and bounded below.
Remark 2.12 If the sequence is not monotone, or if it is not bounded, then it may not be convergent. To see this,

consider for example the sequences ( )∈N with  = (−1) (bounded, but not monotone) or  =  (increasing,
but not bounded above).
From the previous remark we see that not any sequence ( )∈N is convergent.¡ However,¢ we can define two
important “limits” for any sequence, using the previous proposition (by noticing that sup≥  ∈N is a decreasing
sequence, respectively (inf ≥  )∈N is an increasing sequence), as follows.
Definition 2.13 We define the superior limit / lim sup of a sequence ( )∈N by
lim sup  = lim sup  ∈ R ∪ {+∞} 
→∞ ≥
and the inferior limit / lim inf by
lim sup  = lim inf  ∈ R ∪ {−∞} 
→∞ ≥
Remark 2.14 It can be shown that lim inf  ≤ lim sup  for any sequence ( )∈N , and that the sequence ( )∈N
is convergent if and only if we have equality, that is
³
´
lim inf  = lim sup  = lim  
→∞
It can also be shown that lim sup  is the largest limit of a convergent subsequence of the sequence ( )∈N , and
lim inf  is the smallest limit of a convergent subsequence of a the sequence ( )∈N .
Example 2.15 The sequence ( )∈N given by  = (−1) is not convergent, so lim→∞  does not exist.
However, it is easy to see that sup≥  = 1 and inf ≥  = −1 for all  ∈ N, and therefore
lim inf  = −1  +1 = lim sup  
Remark 2.16 (Passing to limit in inequalities) If ( )∈N and ( )∈N are sequences such that  ≤  for
any  ∈ N, it is easy to see that we have
lim inf  ≤ lim inf 
and
lim sup  ≤ lim sup  
If in particular lim→∞  and lim→∞  exists, then we also have
lim  ≤ lim  
→∞
→∞
Remark 2.17 It is important to note that even if    for all  ∈ N (the inequalities are strict), the resulting
inequalities may not be strict, as it can be see by considering  = 1  1 + 1 =  , for which we have lim→∞  =
1 = lim→∞  .
As another sufficient condition for convergence, we have the following:
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Theorem 2.18 (Squeeze theorem) If  ≤  ≤  for all  ∈ N and the sequences ( )∈N and ( )∈N are
convergent to the same limit , then the sequence ( )∈N is also convergent and it has the limit .
Proof. Passing to the limit (inferior limit) in the given inequality, we obtain
lim inf  ≤ lim inf  ≤ lim inf  
and since the sequences  and  are convergent to , we obtain
 ≤ lim inf  ≤ 
that is lim inf  = .
A similar proof shows that lim sup  = , and therefore lim inf  = lim sup  = , which shows that the
sequences  is convergent and lim→∞  = .
The following proposition contains some properties of convergent sequences:
Proposition 2.19 If ( )∈N and ( )∈N are convergent sequences, then:
1. The sequence ( ±  )∈N is also convergent, and we have
lim ( ±  ) = lim  ± lim 
→∞
→∞
→∞
2. The sequence ( ·  )∈N is also convergent, and we have
lim ( ·  ) = lim  · lim 
→∞
3. If  6= 0 and lim→∞  6= 0, then the sequence
→∞
³


´
∈N
→∞
is also convergent, and we have
lim ( ±  ) = lim  ± lim 
→∞
→∞
→∞
4. The sequence (| |)∈N converges to ||.
Proof. 1. If lim→∞  =  and lim→∞  = , then for any   0 we have
|( ±  ) − ( ± )| ≤ | − | + | − |
 2 + 2
= 
for any  ≥  (), since  and  are convergent sequences to , respectively , proving the claim
2. First note that since ( )∈N and ( )∈N are convergent sequences, they are bounded, so there exists   0
such that
| |  | |  
 ∈ N,
and passing to the limit we see that we also have ||  || ≤  .
Given   0 there exists  () such that
| − |  | − | 


2
 ∈ N.
For any  ≥  () we have:
|  − | = | ( − ) + ( − ) |
≤ | | | − | + || | − |


≤ 
+
2
2
 
=
+
2 2
= 
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which shows that the sequence ( ·  )∈N converges to  · .
3. Similar to the previous proof, for 0    ||
2 , we have by writing
¯
¯
¯   ¯
|  −  |
| ( −  ) +  ( − )|
¯
¯
=
¯  −  ¯ =
| |
|2 +  ( − )|
˙ | | | − | + | | | − |
≤
||2 − || | − |
 + 
≤
2 − || 
2 
≤
2
 − || ||
2
4
=

||2
which can be made arbitrarily small (here we need  6= 0), showing that lim→∞  =  .
The following theorem is often useful in exercises for computing limits of the type lim→∞

 :
Theorem 2.20 (Cesaro-Stolz theorem) Let  be a sequence of positive numbers which is increasing to +∞.
Then

+1 − 
lim
= lim

→∞ 
→∞ +1 − 
provided the last limit exists.
−
Proof. We will consider the case when  = lim→∞ +1
is finite (when  = ±∞ the proof is similar).
+1 −
From the definition of the limit it follows that for every   0 there is  such that for any  ≥  , we have :
−
+1 − 
  + 
+1 − 
Because  is strictly increasing we can multiply the last equation with +1 −   0 to get:
( − )(+1 −  )  +1 −   ( + )(+1 −  )
Let  ≥  be a natural number. Summing the above relations for  =    + 1      we get :
( − )

X
=
(+1 −  ) 

X
=
(+1 −  )  ( + )

X
=
(+1 −  )
or equivalent
( − )(+1 −  )  +1 −   ( + )(+1 −  )
Dividing by +1 we obtain:
( − )(1 −

+1


)
−
 ( + )(1 −
)
+1
+1
+1
+1
or equivalent
( − )(1 −
Since  increases to +∞,

+1


+1


)+

 ( + )(1 −
)+

+1
+1
+1
+1
+1
and

+1
tend to 0 as  → ∞, and therefore there exists  such that
( − )(1 − ) −  
for all  ≥ , which shows that
lim
→∞
+1
 ( + )(1 − ) + 
+1
+1
+1 − 
=  = lim

→∞
+1
+1 − 
concluding the proof.
√
As an application, we have the following theorem, useful for computing limits of the type lim→∞   :
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Theorem 2.21 (Root criterion for sequences) Let  be a sequence of positive numbers. Then
lim
→∞
√
+1

 = lim

→∞ 
provided the last limit exists.
Proof. Consider
³
´ 1
√
ln 

 = ln 1
= ln  =




and apply Césaro-Stolz theorem above with  and  replaced by ln  , respectively .
We have
ln
lim
→∞
+1
ln 
ln +1 − ln 
+1
= ln lim

= lim
= lim (ln +1 − ln  ) = lim ln
→∞ ( + 1) − 
→∞
→∞
→∞ 


Since lim→∞
+1

exists, from the Césaro-Stolz theorem we obtain
ln lim
→∞
√
ln 
+1

 = lim

= ln lim
→∞ 
→∞ 
or equivalent
lim
→∞
√
+1

 = lim

→∞ 
concluding the proof.
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2.1
Exercises
1. Show that if ( )≥1 converges, then (| |)≥1 also converges. Is the converse true? (either prove it or give
a counterexample).
¢
¡√
2. Calculate lim→∞ 2 +  − 
√
p
√
3. If 1 = 2 and +1 = 2 +  ,  ≥ 1, show that the sequence ( )≥1 converges and find its limit.
4. Compute the limits of the following sequences:
√
√
(a)  =  + 1 − 
(b)  =
√
√
+1− 

´
,  ≥ 1, where   0 is given number. Show that ( )≥1 is a
√
decreasing, bounded sequence and its limit is lim→∞  = .
5. Let 1 
√
 and +1 =
1
2
³
 +


6. Find the limit of the following sequences:
 =
3 + 1
22 + 5
32 + 4
2 − 1
 =
 =
42 + 2 − 1
52 + 10
7. Find the limit of the following sequences:
 =
4
3 + 1
 =
2 + 3
5 − 1
 =
5 − 2
3 + 1
8. Find the limit of the following sequences:
√
√
√
√
−1− +2
 = √ 2 − 3 +√2 − 2 + 2 + 1  = 2√
 =  − 2 − 3 + 2
 = 4 + 3 − 4 − 2
9. Find the limit of the following sequences:
µ
¶2
3
 = 2 +

µ
¶
4
 = 5 +

10. Find the limit of the following sequences:
+2
 = √
3 + 4
2 − 2 − 1
 = √
44 − 33
2 − 3
 = √
23 + 1
p
 = 3 − 2 + 1
11. Find the limit of the following sequences:
 =
µ
4
1+
−2
¶52 +2
 =
µ
−2
+3
¶
2 +2
3 +4
 =
µ
2 − 3 + 2
2 + 4 + 4
¶ 2 +2

12. Find the limit of the following sequences:
3 + 4 · 5 − 6 · 7
2 + 3 · 4 + 5 · 7
¡
¢ 3 + 4
 = 1 + 3 + 32 +    + 3
5
 =
13. Find the limit of the following sequences:
p
√
¡√
¢
 = 3 + 2  + 2 −  − 1
2 + 3 · 4 − 5 · 6
µ4 − 2 · 5 + 7
¶
1
1 (−1) + 4
1
 = 1 + + 2    + 
3 3
3
2 · 4
 =
 =
21
³p
´
p
p
2 + 1
2 + 2 − 2 − 3
14. Find the limit of the following sequences:
s

 =
( + 1)!
(2 + 1)! · 3
 =
s

(( + 1)!)2
(2 + 1)! · 3
15. Find the limit of the following sequences:
 =
1+
1
2
+  +

1+
1

 =
√1
2
+  +
√

√1

 =
1
ln 2
+
1
ln 3
+  +
2
1 + 2 +    + 
where  ∈ N is a natural number.
→∞
+1
16. Find the limit of the sequence lim
Additional exercises
17. Find the given limits
(a) lim→∞
√
2+√
2− 
¡√
√ ¢
+1− 
¡√
√ ¢
(c) lim→∞ 3  + 1 − 3 
P
(+)10
(d) lim→∞ 100
=1 10 +1010
(b) lim→∞
s r
q
√
(e) lim→∞  , where  = 2 2 2    2
{z
}
|
(f)
(g)
(h)
 radicals
¢
1
2
−1
lim→∞ 2 + 2 +    + 2
lim→∞ (+1)(+2)(+3)
3
lim→∞ 1+2+3++
2
¡
(i) lim→∞
(j) lim→∞
(k) lim→∞
³
1+3+5+7++(2−1)
+1
−
2+1
2
+(−1)
−(−1)
´
2+1 +3+1
2 +3
2·3 +(−3)
4
3 +
lim→∞ 4 +5 , where    0
¡
¢
lim→∞ 12 + 14 + 18 +    + 21
(l) lim→∞
(m)
(n)
³
(o) lim→∞ 1 −
1
3
+
1
9
−
1
27
+  +
are given numbers.
(−1)−1
3−1
´
18. Find the given limits (Hint: squeeze theorem).
(a) lim→∞
(−1)

cos2 

!
lim→∞ sin
2 +1
(b) lim→∞
(c)
(d) lim→∞
(e) lim→∞
(f) lim→∞
(g) lim→∞
(h) lim→∞
2
!

! , where  ∈ R is a given number

√
, where  ∈ R is a given number
!
P sin 
=1 2 +
P
√ 1
=1 2 +
22
1
ln 
(j)
P
2 +
=1 3 +
P
cos 
lim→∞ =1 2 ++
2
(i) lim→∞
P []+[22 ]+[32 ]++[2 ]
(k) lim→∞ =1
, where [] denotes the integer part of  (recall that [] ≤   []+1
2
for any  ∈ R)
19. Are the sequences with general term  = 1 + (−1)
Why/why not?
+1
10 ,
respectively  = 1 + (−1)
+1
102 ,
convergent?
20. Show that the given sequences ( )≥1 are convergent.
(a)  =
(b)  =
(c)  =
(d)  =
P
1
=1 2 ,
P
 ≥ 1.
sin(2 )
,
=1
2
P cos(!)
=1 (+1) ,
P
cos 
=1 (+1) ,
 ≥ 1.
 ≥ 1
 ≥ 1, where ( )≥1 is a given sequence of real numbers.
Hint: convergent sequence ⇐⇒ Cauchy sequence
21. Show that the given sequences ( )≥1 converge, then find the corresponding limit.
(a) 1 = 1, +1 =
(b) 1 = 1, +1 =
(c) 1 = 1, +1 =

1+ ,
2
≥1
1+ ,  ≥ 1

1+ ,  ≥ 1,
where    0 are given numbers.
¢
¡ 2
 +  ,  ≥ 1, where  ∈ [0 1] is a given number.
(d) 1 = 0, +1 =
√
√
(e) 1 = , +1 =  +  ,  ≥ 1, where   0 is a given number.
1
2
22. Compute the limits of the given sequences ( )≥1 .
(a)  =
(b)  =
(c)  =
(d)  =
(e)  =
(f)  =
(g)  =
(h)  =
(i)  =
√ √
√
1+ 2+ √
3++ 
 
1+ √12 + √13 ++ √1

√
√
√
1+22 2+32 3 3++2  
(+1)(+2)
ln ·ln !

ln 

1 +2 +3 ++

− +1

√
P  +
1
√
=1  + , where    

1

³
√

√

(k)  = 
(j)  =
(l)  =
(m)  =
 0 are given numbers
21 +22 ++2

1
1+1
, where ( )≥1 is a sequence with lim→∞  = 5
´
2

+ 1+
+



+
1+ , where ( )≥1 is a sequence with lim→∞  = 0
2

√

!
q

33 (!)3
(3)!
√

1 + 2 + 3 +    +  , where  ∈ R is a given number.Does the answer depend on ?
√
(o)  =   where   0 is a given number
qP

(p)  =  =1 1
(n)  =
23
√

 +  , where    0 are given numbers.
√
(r)  =   +  +  , where     0 are given numbers. Can you generalize?
(q)  =
23. For the given sequences ( )≥1 , first find a simplified formula for the general term  , then find its limit.
(a)  =
(b)  =
(c)  =
(d)  =
P
1
=1 (+1)
P
2+1
=1 2 (+1)2

(+1)2
 =1
(+1)(+2)
=1

2−
=0

−
=0 3
23 −1
23 +1
33 −1
33 +1
3
·    · 3 −1
+1
³
´
P
2
(f)  = =2 log3 1 − (+1)
P
(g)  = =1 (2−1)2(2+1)2
P
2
+−1
(h)  = =1 (+2)!
(e)  =
·
24. Consider the sequence ( )≥0 defined by 0 = 1 and +1 =  + 1 . Does the sequence ( )≥0 have a
limit? Is the sequence ( )≥0 convergent?
³
´
25. Consider the sequence ( )≥0 defined by 0 = 2 and +1 = 12  + 1 . Does the sequence ( )≥0 have
a limit? Is the sequence ( )≥0 convergent?
³
´
√
√
√+1+√+2 , where     0 are given numbers.
26. Find lim→∞  ln +
+ +2+ +3
27. Find lim→∞
33

(3+ 1 )3
¡
28. Consider the sequences ( )≥1 and ( )≥1 defined by  = 1 +
(a) Find a simplified formula for 
¢
 

and  =
P
1
=1 (2−1)(2+1) ,
≥1
(b) Find lim→∞ 
(c) Find lim→∞
+1

29. Consider the sequence ( )≥1 defined by  =
(a) Find  = lim→∞ 
(b) Find lim→∞  , where  =
30. Find the indicated limits
¢
¡
P
(a) lim→∞ =2 ln 1 − 12
³ P
´
1
(b) lim→∞ 43 =1 (+2)
(c) lim→∞
(d) lim→∞
P
=1  ,
³
2 −+1
2 +1
++1
´ 2+1
,  ≥ 1, where  ∈ R is a given number
 ≥ 1.
√
√
√
3++ 
1+2 2+3 √
2
 
q

(!)
()! ,
where  ∈ N∗ is positive natural number.
31. Consider the sequence ( )≥2 , where  =
(a) Does the sequence ( )≥2 converge?
p

(−1) .
(b) What is wrong with the following “proof”? “By the root criterion for sequences, we have lim→∞
+1
lim→∞ (−1)
(−1) = lim→∞ −1 = −1, so the sequence ( )≥2 converges to −1”
24
p

(−1) =