Download The Limit of a Sequence

The Limit of a Sequence (Brief Summary)
1. Definition. A real number L is a limit
of a sequence of real numbers if every
open interval containing L contains all
but a finite number of terms of the sequence.
2. Claim. A sequence can have at most
one limit.
Proof: Suppose L and L0 be limits of
a sequence. Then there exists an open
interval I containing L and an open interval I 0 containing L0 and both of these
intervals contain all but a finite number of terms of the sequence. If L 6= L0 ,
I and I 0 can be chosen to be disjoint
which (exercise) leads to a contradiction. Thus, it must be that L = L0 .
3. Notation. If L is the limit of a sequence
(an ) we write
lim an = L
n→∞
and we say that the sequence converges
to L or that the sequence is convergent.
A sequence with no limit is called divergent.
4. Remark. The definition of limit gives
precise meaning to the rather vague
phrase “an approaches L as n approaches ∞”. This statement does not
serve as the definition of limit because
neither the nature of the “approach”
nor the concept of ∞ are explained.
5. Lemma. Let I is any open interval containing L, then within I can be found
an open interval of the form |x − L| < 1
1
6. Theorem. Let (an ) be a sequence. The
following are equivalent,
(a) lim an = L
n→∞
(b) For every > 0, all but a finite
number of terms of the sequence
are contained in the interval with
center L and radius .
(c) ∀ > 0, ∃N such that
n > N =⇒ |an − L| < 7. Example: Let (an ) be the sequence
given by
n
.
an =
n+1
Show that lim an = 1.
n→∞
Solution: Let > 0 be arbitrary. We
must find a natural number N such
that2
d(an , 1) < (1)
whenever n > N . Now
n
1
d(an , 1) = − 1 =
.
n+1
n+1
Moreover, since IN is unbounded above
there exists a natural number N such
that N > 1. It follows that:
1
⇒ n > 1
1
⇒ <
n
1
⇒
<
n+1
⇒ d(an , 1) < n>N ⇒n>
Philip Pennance http://pennance.us - Version: April 10, 2017.
Notice that the value of N will depend on the value of . The smaller the value of the larger N must
be in order that the open interval determined by equation (1) contains all terms after the N ’th. Figures 3
and 4 suggest that that if = 0.2 then N = 5 works and if = 0.1 then N = 9 suffices.
2
N such that |an − 5| < whenever
n > N. Let N = 24 (any other natural
number will do). Since |an − 5| = 0,
for all n ∈ IN it is clearly true that
|an − 5| < whenever n > 24. It follows that lim an = 5.
Since N has the required property, we
conclude that
n
=1
n→∞ n + 1
lim
8. Example. Let (an ) be the constant sequence given by an = 5, n ∈ IN. Show
that lim an = 5.
n→∞
9. More generally, if an = c for all n ∈ IN
then lim an = c.
n→∞
Solution. Let > 0. It is easy to find
n→∞
a1
a2
a3
a4
···
0
.1
.2
.3
.4
?
s
.5
.6
?
s
.7
.8
6
s
0.5
0
1
s
2
.9
?
s -
1
n
n+1
s
s
s
s
s
s
s
s
3
4
5
6
7
8
9
10
Figure 2. Graph of the sequence an =
L + 0.2 = 1.2
L = 1.0
L − 0.2 = 0.8
···
?
s s s s ssss?
Figure 1. Initial Terms of the Sequence an =
1
a50
-
n
n+1
6
s
0.5
0
1
s
2
s
s
s
s
s
s
s
s
3
4
5
6
7
8
9
10
Figure 3. n > 5 ⇒ |an − 1| < 0.2.
2
-
6
L + 0.1 = 1.1
L = 1.0
s
s
0.5
0
1
2
s
s
s
s
s
s
s
s
3
4
5
6
7
8
9
10
L − 0.1 = 0.9
-
Figure 4. n > 9 ⇒ |an − 1| < 0.1.
12. Theorem.
Let (an ) and (bn ) be sequences.
lim an = L and lim bn = M then:
10. Example: Let (an ) be the sequence
1
, n > 1. Show usgiven by an =
n−1
ing the definition of the limit of a sequence that
n→∞
If
n→∞
(a) lim (an ± bn ) = L ± M
n→∞
lim an = 0.
(b) lim (an · bn ) = L · M
n→∞
n→∞
(c) If M 6= 0 and bn 6= 0 for all n then
Solution. Let > 0 Notice (exercise)
1
< n2 for all n > 2. Since IN is
that n−1
unbounded, there exists a natural number N > 1 such that N > 2. It follows
that:
L
an
.
=
n→∞ bn
M
lim
13. Example. Find the limit of the sequence (an ) given by
n > N ⇒ n > 2
2
⇒ <
n
Since
1
n−1
<
2
,
n
an = 3/n
Solution. Notice an = xn · yn where
xn = 1/n and yn = 3, n ∈ IN. Now,
it is easy to show that lim xn = 0 and
it follows that
1
<
n − 1
1
⇒ − 0 < n−1
⇒ d(an , 0) < n→∞
n>N ⇒
lim yn = 3. Using part (b) of the above
n→∞
theorem yields
lim an = 0 · 3 = 0
n→∞
14. Example. Find the limit of the sequence (an ) given by
which proves that
1
lim
=1
n→∞ n − 1
an =
1
.
n2
Solution. Let xn = 1/n. Then
11. The next theorem reduces the calculation of complex limits to simpler ones.
an = x n · x n
3
Since lim xn = 0.
17. Definition. A sequence (an ) is bounded
if there exists a number M such that
|an | ≤ M for all n ∈ IN. A sequence which is not bounded is called
unbounded.
n→∞
lim an = lim xn · lim xn
n→∞
n→∞
n→∞
=0·0
= 0.
18. Example. Consider the sequence an =
n
. Since |an | ≤ 1 for all n ∈ IN,
n+1
this sequence is bounded.
15. Using mathematical induction (exercise) it is now easy to show that
1
=0
n→∞ np
lim
19. Claim.
A convergent sequence is
bounded.
Proof. Let (an ) be a convergent sequence with limit L. Then there exists a number N such the open interval
(L − 1, L + 1) contains all terms an with
n > N . It follows that all terms of an
are less than or equal in absolute value
to the maximum M of the set
for any natural number p > 0.
16. Example.
Find the limit of the sen2 + 3n
.
quence xn =
3 + n2
Solution.
n2 3n
+ 2
2
n
xn = n
3
n2
+
n2 n2
3
1+
n
=
3
+1
n2
{|a1 |, |a2 |, . . . , |aN |, |L + 1|, |L − 1|}
(Exercise: Find a counterexample to
show that the converse of this result is
false.)
But
20. Corollary. The sequence an = n has no
limit.
Proof. Notice that the sequence (an ) is
unbounded. If it had a limit it would
be bounded.
3
3
lim 1 +
= lim 1 + lim
n→∞
n→∞
n→∞ n
n
=1+0
=1
21. Sandwich Theorem
Let (an ), (bn ) and (cn ) be sequences. If
an ≤ bn ≤ cn for all n ∈ IN and the
seqences (an ) and (cn ) have the same
limit L then limn→∞ bn = L
cos n 22. Example. Show lim =0
n→∞
n
Solution. Notice that for all n ≥ 1
cos n 1
0≤
≤
n
n
and
lim
n→∞
3
+1
n2
3
+ lim 1
n→∞ n2
n→∞
=0+1
=1
= lim
Hence
3
lim 1 +
n→∞
n
lim xn =
n→∞
3
lim
+1 .
n→∞
n2
= 1.
The result follows by applying the sandwich theorem.
4
(b) non-decreasing if for all n, m ∈ IN
23. Claim. The following are equivalent:
(a) lim an = 0.
n ≤ m ⇒ xn ≤ x m
n→∞
(b) lim |an | = 0.
(c) decreasing if for all n, m ∈ IN
n→∞
Proof. Exercise
Hint: (Notice that
n < m ⇒ xn > xm
(d) non-increasing if for all n, m ∈ IN
d(|an |, 0) = d(an , 0)
24. Corollary. limn→∞
n ≤ m ⇒ xn ≥ xm
cos n
=0
n
Warning: As shown in the Venn diagram, the terms “non-increasing” and
“increasing” are not mutually exclusive.
25. Lemma. (Bernoull’s Inequality)
If x > −1 then
(1 + x)n ≥ (1 + nx)
Proof. By induction (Exercise).
26. Claim. If a > 1 then an is unbounded
above.
Proof. If a > 1 then a = 1 + x where
x = a − 1. Notice x > 0. By Bernoulli’s
inequality
Increasing
Decreasing
inc
A
C
dec
Non-decreasing
an = (1 + x)n > 1 + nx
Non-increasing
29. Dedekind (Completeness) Axiom.
Every non-empty subset of IR which is
bounded above has a supremum (least
upper bound).
Since IN is unbounded above, given M
arbitrary, there exists N ∈ IN such that
for all n > N , nx > M − 1. Hence,
for all n > N an > M proving that the
sequence (an ) is unbounded above.
30. Claim. A sequence α = (an ) of real
numbers which is both non-decreasing
bounded converges to the least upper
bound of its set of terms.
Proof. Let L = sup{an : n ∈ IN} and
> 0. Then there exists N such that
n
27. Claim. If 0 < a < 1 then lim a = 0
n→∞
1
a
Proof. Since > 1, it follows by the
n
previous claim that the sequence a1
is unbounded. It follows that if > 0
there
N such that for all n > N ,
exists
1 n
1
> . This implies that|an − 0| < a
for all n > N . Hence lim an = 0.
L − < aN ≤ L
n→∞
(otherwise L − would be a smaller
bound). Since α is non-decreasing
28. Definition. A sequence (xn ) in IR is:
(a) increasing if for all n, m ∈ IN
n ≥ N ⇒ L − < an ≤ L
n < m ⇒ xn < xm
Hence α converges to L.
5
31. Corollary. A non-decreasing sequence
α = (an ) of real numbers either converges or is unbounded above.
.
36. Claim. Cauchy sequences are bounded.
32. Definition. Let (xn ) be a real sequence.
A natural number m is called:
Proof. Let > 0. There exists N such
that for m, n ≥ N , |am −an | < . Then,
by the triangle inequality
(a) A peak point of (xn ) if xm ≥ xn for
all n ≥ m.
|am | − |an | ≤ |am − an |
<
(b) A p-blocker if m > p and xm ≥ xp
33. Bolzano Weierstrass Theorem.
A bounded sequence has a convergent
subsequence.
Proof. (See Spivak).
Let m1 < m2 , < . . . , be the set of peak
points. If this set is infinite then the
bounded non increasing subsequence
for all m, n ≥ N. Taking n = N . Then
|am | − |aN | < for all m ≥ N . It follows easly that
(am ) is bounded by
± max {|a0 |, |a1 |, . . . , |aN −1 |, |aN |, + |aN |}
xm1 ≥ xm2 ≥ xmk . . .
must converge (to its infimum). If, on
the other hand, the set of peak points
is finite, there exists a last peak point
m in which case there exists a convergent subsequence of m-blockers defined
recursively as follows: let q1 = m + 1
and, for each k, let qk+1 be the minimum blocker of qk . Then the subsequence
37. Claim [Uses BW]. Every Cauchy sequence in IR converges.
Proof. Let (an ) be a Cauchy sequence
of real numbers. Then (an ) is bounded
so by BW has a convergent subsequence
(ank ). Let L be the limit of this subsequence. Then there exist N such that
nk > N ⇒ |ank − L| < /2
xq1 ≤ xq2 ≤ xqk . . .
and there exists M such that
is non decreasing and bounded below so
converges.
n, nk > M ⇒ |an − ank | < /2
34. Exercise. In the proof of BW above, explain why the minimum at each stage
exists.
Then, for all n > max{N, M }
|an − L| ≤ |an − ank | + |ank − L|
<
35. Definition. A real sequence (an ) is
Cauchy if for every > 0 there exists
N such that
38. Corollary. Dedekind completeness implies Cauchy completeness
n, m > N ⇒ |an − am | < 6