Download MATH 201: LIMITS 1. Sequences Definition 1 (Sequences). A

MATH 201: LIMITS
1. Sequences
Definition 1 (Sequences). A sequence an is a real-valued function
(1)
T → R; n 7→ an
whose domain T is an inductive subset of the set N of natural numbers.
Definition 2 (Terms of a sequence). The real-number values an of
a sequence (1) are known as the terms of the sequence.
Example 3 (Powers of 2). The sequence 2n of powers of 2 is
1, 2, 22 = 4, 23 = 8, . . .
In this case, the domain is the inductive subset T = N of N consisting
of the natural numbers (starting at 0 to obtain 1 = 20 ).
Example 4 (The harmonic sequence). The harmonic sequence n1
is the sequence
1 1 1
1, , , , . . .
2 3 4
of reciprocals of positive integers. In this case, the domain T is the set
of positive integers.
In Examples 3 and 4, the sequences were given directly by listing
their terms as 2n and n1 . Sometimes, sequences are defined recursively.
Definition 5 (Geometric sequences). Let a and r be real numbers.
Then the geometric sequence an with initial term a and ratio r is
defined recursively by
(
a0 = a and
(2)
ak+1 = ak r
for natural numbers k.
Example 6 (A savings account). When Sam was born, the grandparents put $10, 000 into a savings account returning 4% interest per
year. Thus the dollar amount of money an in the account on Sam’s
n-th birthday is given by
a0 = 10000 and
ak+1 = 1.04ak
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MATH 201: LIMITS
for natural numbers k. This is a geometric sequence with initial term
10, 000 and ratio 1.04. On Sam’s 21-st birthday, the account is worth
$22, 788.
Theorem 7 (Terms of a geometric sequence). For each natural
number n, the term an of the geometric sequence (2) is given by
an = arn .
Proof. By induction on n, using the recursive definition.
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2. Limits of sequences
The sequence of powers of 2 (Example 3) is
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, . . .
As n increases towards infinity, the terms of the sequence increase
rapidly towards infinity. On the other hand, the terms
1 1 1 1 1 1
1, , , , , , , . . .
2 3 4 5 6 7
of the harmonic sequence (Example 4) gradually decline towords 0 as
n increases towards infinity.
Definition 8 (Tending to infinity). A sequence an is said to tend
to infinity (as n tends to infinity), or to have infinity as its limit, if the
following is true:
(3)
∀ K ∈ R , ∃ N ∈ N . ∀ n ≥ N , an ≥ K .
Symbolically, we write
lim an = ∞
n→∞
to state that the sequence an tends to infinity (as n tends to infinity).
Note that the third quantifier phrase “∀ n ≥ N ,” in (3) may be
rewritten as
∀ n ∈ {m ∈ N | m ≥ N } ,
using set-builder notation, or as
∀ n ∈ {N, N + 1, N + 2, N + 3, . . . } ,
using the roster method.
The property (3) states that, however high the bar K is set, we can
always go far enough along the list of terms of the sequence (specifically,
as far as the term aN ), so that from that point on, all the terms an of
the sequence are above the bar (Figure 1).
MATH 201: LIMITS
3
6
an
aN +1
aN
t
t
K
t
aN +2
t
-
N
N +1 N +2
...
n
Figure 1. Sequence an tending to infinity.
Proposition 9.
lim n2 = ∞ .
n→∞
Proof. Suppose that a real number K is given. We must show that
there is a natural number N so that n2 ≥ K for all n ≥ N .
Case 1: K < 0.
If K is negative, we may choose N = 0 to guarantee n2 ≥ K for n ≥ N ,
since all the terms n2 are nonnegative.
Case 2: K ≥ 0.
√
In this case, round
K up to the next integer N . (For example,
√ if
√
K = 3, then 3 = 1.732 . . . rounds up to N = 2.) Note N ≥ K.
Then n ≥ N implies
√
n≥N ≥ K,
√
so that n ≥ K and
√
n2 ≥ ( K)2 = K .
Thus n2 ≥ K for all n ≥ N , as required.
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To show that a sequence an does not tend to infinity, we have to
prove the negation
(4)
∃ K ∈ R . ∀ N ∈ N , ∃ n ≥ N . an < K
of the property (3). As before, the third quantifier phrase “∃ n ≥ N .”
in (3) may be rewritten as
∃ n ∈ {m ∈ N | m ≥ N } .
using set-builder notation, or as
∃ n ∈ {N, N + 1, N + 2, N + 3, . . . } .
using the roster method.
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MATH 201: LIMITS
Proposition 10. The sequence 2n cos nπ does not tend to infinity.
Proof. Consider K = 0. For a given natural number N , choose an odd
integer n greater than N . Then 2n cos nπ = −2n < 0 = K.
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Warning! Although the symbolic equation
lim an = ∞
n→∞
is used to express the fact that a sequence an tends to infinity, we
cannot use the symbolic equation
(5)
lim an 6= ∞
n→∞
to express the fact that a sequence an does not tend to infinity. (This
is one case in which “x 6= y” is not the negation of “x = y”!)
If the inequality (5) is claimed to hold, then part of the claim is
that the left hand side of (5) exists (as a limit in the general sense of
Definition 18 below). However, saying that a certain sequence has a
limit, which happens to differ from infinity, is not the same as saying
that the sequence does not tend to infinity. For example, a sequence
may well have no limit at all, finite or infinite.
3. Limits of geometric sequences
Theorem 11 (Binomial Theorem). For a real number x and natural
number n, we have
n µ ¶
X
n l
n
(1 + x) =
x
l
l=0
µ ¶
n 2
x + · · · + xn .
= 1 + nx +
2
The Binomial Theorem is proved by induction on n.
Corollary 12 (Linear/Binomial Inequality). For a positive real
number x and natural number n, we have
1 + nx ≤ (1 + x)n .
Proof. If the real number x is positive, then
µ ¶
n 2
(6)
0≤
x + · · · + xn
2
since the powers of x and the binomial coefficients are positive. Note
that the right-hand side of the inequality (6) is 0 for n < 2.
Now consider the equality
(7)
1 + nx = 1 + nx .
MATH 201: LIMITS
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Adding the equality (7) to the inequality (6), we obtain
µ ¶
n 2
(8)
1 + nx ≤ 1 + nx +
x + · · · + xn .
2
Finally, using the Binomial Theorem, we may rewrite the right-hand
side of the inequality (8) to obtain
1 + nx ≤ (1 + x)n ,
which is the Linear/Binomial Inequality.
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Theorem 13 (Limit of a geometric sequence). Suppose that an is
a geometric sequence with positive initial term a and ratio r > 1. Then
an tends to infinity as n tends to infinity.
Proof. Let r = 1 + x. Note that the real number x is positive. By
Theorem 7, we have
an = arn = a(1 + x)n .
Given a real number K, choose a natural number N with
N≥
K −a
.
ax
Thus aN x ≥ K − a and a(1 + N x) = a + aN x ≥ K. Now for n ≥ N ,
the Linear/Binomial Inequality gives
an = a(1 + x)n ≥ a(1 + nx) ≥ a(1 + N x) ≥ K ,
as required.
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Theorem 13 shows that the powers of 2 (Example 3) tend to infinity:
Set the initial term a = 1 and ratio r = 2 in the theorem.
4. Finite limits.
Definition 14 (Tending to zero). A sequence an is said to tend to
zero (as n tends to infinity), or to have zero as its limit, if the following
is true:
(9)
∀ ε > 0 , ∃ N ∈ N . ∀ n ≥ N , |an | < ε .
Symbolically, we write
lim an = 0
n→∞
to state that the sequence an tends to zero (as n tends to infinity).
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MATH 201: LIMITS
ε
6
aN
t
aN +2
t
N
N +1 N +2
an
t
...
-
n
aN +1
t
−ε
Figure 2. Sequence an tending to zero.
The use of the Greek letter ε (epsilon) for the positive real number in
the first quantifier clause of (9) is traditional. Think of ε as a tolerance,
like the tolerance required of machine parts that are supposed to fit
together. Intuitively, we are claiming to be able to get the sequence
terms an as close to zero as anybody would require. If they ask us to
make the terms stay within a distance ε of zero, we guarantee that this
will happen for all an once the index n is no less than our specified
number N .
Proposition 15. The harmonic sequence
1
n
tends to zero.
Proof. For a real number ε > 0, choose a natural number N with
N>
1
.
ε
Then for n ≥ N , we have
n≥N >
1
ε
or n >
which gives
1
<ε
n
after inversion. Finally, note that
¯ ¯
¯1¯ 1
¯ ¯=
¯n¯ n
1
,
ε
MATH 201: LIMITS
since n > 0, so we obtain
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¯ ¯
¯1¯
¯ ¯<ε
¯n¯
for n ≥ N , as required.
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To show that a sequence does not tend to zero, we must establish
the negation
(10)
∃ ε > 0 . ∀ N ∈ N , ∃ n ≥ N . |an | ≥ ε .
of the condition (9). Intuitively, the condition (10) states that there is
some tolerance ε such that infinitely many terms of the sequence miss
zero by at least that tolerance.
Proposition 16. The sequence
n−1
n+1
does not tend to zero as n tends to infinity.
an =
Proof. Let ε = 21 . Given a natural number N , choose n to be the
maximum of N and 3. In particular, n ≥ 3. Now
n≥3
⇒ 2n ≥ n + 3
⇒ 2(n − 1) = 2n − 2 ≥ n + 1
1
⇒ n − 1 ≥ (n + 1)
2
n−1
1
⇒
≥ .
n+1
2
Since
in this case, we have
n−1
≥0
n+1
¯
¯
¯n − 1¯ n − 1
¯
¯
¯n + 1¯ = n + 1 ,
so
n−1
1
≥
n+1
2
means
¯
¯
¯n − 1¯
¯
¯
¯n + 1¯ ≥ ε
for the chosen n ≥ N , as required.
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MATH 201: LIMITS
Note that the sequence an of Proposition 16 does not tend to zero,
despite the fact that its term a1 is actually zero.
The final result compares with Theorem 13.
Theorem 17 (Limit of a geometric sequence). Suppose that an is
a geometric sequence with ratio r satisfying |r| < 1. Then an tends to
zero as n tends to infinity.
Proof. Let a be the initial term a0 of the sequence. By Theorem 7, we
have
an = arn
for all natural numbers n. If the ratio r is zero, all the terms after a0
are zero. Given ε > 0, the inequality |an | < ε holds for all n ≥ 1, so
the sequence also tends to zero in this case.
Now suppose that |r| 6= 0. Since |r| < 1, we have
1
> 1,
|r|
so
1
=1+x
|r|
for a certain positive real number x. Given ε > 0, choose a natural
number N with
|a| 1
N>
− .
εx x
Then for n ≥ N , we have
|a| 1
−
εx x
|a|
⇒ nx >
−1
ε
|a|
⇒ 1 + nx >
.
ε
n>
Using the Linear/Binomial Inequality, we obtain
|a|
1
= (1 + x)n ≥ 1 + nx >
,
n
|r|
ε
so that
|an | = |arn | = |a| · |r|n < ε ,
as required to show that an tends to zero.
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MATH 201: LIMITS
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Definition 18 (General limits). Let L be a real number. A sequence
an is said to tend to L or to have L as its limit if the sequence an − L
tends to zero. Symbolically, we write
lim an = L
n→∞
to mean that
lim (an − L) = 0 .
n→∞
General limits are studied in advanced calculus and analysis.