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Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
8.1
Sequences
We start this section by introducing the concept of a sequence and study its
convergence.
Convergence of Sequences.
An infinite sequence is a list of numbers a1 , a2 , a3 , · · · .
Example 8.1.1
(1) 1, 12 , 13 , · · · .
(2) 0, 2, 0, 2, 0, · · · .
Formally, a sequence is a function f with domain the set of nonnegative
integers IN. The image of n ∈ IN is denoted by f (n) = an . We represent a
th term of the sesequence by the notation {an }∞
n=1 and we call an the n
quence. For example, the nth term of the sequence (1) above is an = n1 , n ≥ 1
while the nth term of (2) is an = 1 + (−1)n .
We say that a sequence {an }∞
n=1 converges to a number L if and only
if an approaches L as n gets larger and larger, i.e., if the difference |an − L|
can be made as small as we wish by taking n large enough. Using the -δ
argument, if > 0 is given then we can find a positive integer N such that
|an − L| < f or n ≥ N.
We write
lim an = L.
n→∞
Figure 8.1.1 illustrates the definition of convergence of a sequence.
Figure 8.1.1
If a sequence does not converge then we say that it diverges.
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Example 8.1.2
2 +5
3n2 +5 ∞
(1) Since lim n3n
2 +n+1 = 3, the sequence { n2 +n+1 }n=1 converges to 3.
n→∞
(2) Since (−1)n alternates between −1 and 1, the sequence {1 + (−1)n }∞
n=1
alternates between 0 and 2 and so it diverges.
The following theorem shows the connection between the convergence of a
function and the convergence of a sequence.(See Figure 8.1.2). This basically
allows us to replace limits of sequences with limits of functions. In particular,
this is useful for using L’Hôpital’s rule in computing limits of sequences.
Theorem 8.1.1
If lim f (x) = L and an = f (n) then lim an = L.
x→∞
n→∞
Figure 8.1.2
Proof.
Let > 0 be given. Since lim f (x) = L, we can find a δ > 0 such that
x→∞
|f (x) − L| < if |x| ≥ δ.
Let N = (integer part of δ) + 1. Then N is a positive integer greater than
or equal to δ. Thus, if n ≥ N then
|an − L| = |f (n) − L| < .
This shows that lim an = L
n→∞
Example 8.1.3
(n+1)
(a) Find lim ln n+1
.
n→∞
(b) Find lim
n→∞
sin n
√
.
n+1
2
Solution.
(a) Applying L’Hôpital’s rule we have
lim
n→∞
ln (n+1)
n+1
=
=
lim
x→∞
lim
lim
x→∞
√1
x+1
1
x+1
1
x→∞
1
≤
(b) Since −1 ≤ sin n ≤ 1 we obtain − √n+1
ln (x+1)
x+1
= 0.
sin n
√
n+1
≤
√1 .
n+1
But lim
n→∞
√1
n+1
=
= 0. By the squeeze rule we have
sin n
=0
lim √
n+1
n→∞
We next discuss some of the important properties of sequences.
Theorem 8.1.2 (Convergence Properties of Sequences)
∞
Let {an }∞
n=1 and {bn }n=1 be two sequences such as lim an = L and lim bn =
L0 where L and L0 are finite numbers. Then
(a) lim (an + bn ) = lim an + lim bn = L + L0 .
n→∞
n→∞
n→∞
n→∞
n→∞
(b) lim kan = k lim an = kL, where k is a constant.
n→∞
n→∞
(c) lim an bn = ( lim an )( lim bn ) = LL0 .
n→∞
(d)
lim an
n→∞ bn
=
n→∞
lim an
n→∞
lim bn
n→∞
=
n→∞
L
L0
provided that lim bn 6= 0.
n→∞
(e) Squeeze Principle: If an ≤ bn ≤ cn for all n ≥ N0 and if
lim cn = L then lim bn = L.
n→∞
lim an =
n→∞
n→∞
(f) If lim |an | = 0 then lim an = 0.
n→∞
n→∞
Proof.
We will prove (a) and leave the rest as an exercise. Let > 0 be given. Then
there exist positive integers N1 and N2 such that
|an − L| <
f or n ≥ N1
2
and
f or n ≥ N2 .
2
Let N = N1 + N2 . Then for n ≥ N we have
|bn − L0 | <
|(an + bn ) − (L + L0 )| ≤ |an − L| + |bn − L0 | <
3
+ = .
2 2
Example 8.1.4
Show that the sequence with nth term an =
Solution.
Since
(−1)n
n
converges to 0.
(−1)n = lim 1 = 0,
lim n→∞
n n→∞ n
By (f) of the above theorem, we have
(−1)n
=0
n→∞
n
lim
In order to discuss the next result we need the following definitions: We say
that a sequence {an }∞
n=1 is bounded from below if there is a number M
such that M ≤ an for all n. It is said to be bounded from above if an ≤ M
for all n. We say that a sequence is bounded if there are two numbers K
and M such that K ≤ an ≤ M for all n.
The next result shows that convergent sequences are always bounded.
Theorem 8.1.3
Suppose that lim an = L. Then there exists a positive number M such
n→∞
that −M ≤ an ≤ M for all n. That is the sequence is bounded.
Proof.
Let = 1. Then there exists N > 0 such that |an − L| < 1 whenever n ≥ N.
This implies that L − 1 < an < L + 1 for all n ≥ N. As a result of that we
have |an | < L + 1 for all n ≥ N. Let M = max{|a1 |, |a2 |, · · · , |aN −1 |, L + 1}.
Then |an | ≤ M for all n, i.e. −M ≤ an ≤ M for all n. This establishes a
proof of the theorem
Now, we discuss a very useful theorem that establishes the convergence
of a given sequence (without, however, revealing the limit of the sequence).
But first we introduce the following definitions: We say that a sequence
{an }∞
n=1 is increasing if an ≤ am whenever n ≤ m. A sequence is said to
be decreasing if am ≤ an whenever n ≥ m.
Theorem 8.1.4 (Monotone Convergence Theorem)
An increasing sequence that is bounded from above is always convergent. A
decreasing sequence that is bounded from below is always convergent.
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Proof.
We will prove the first part and leave the second part for the reader. Since
{an }∞
n=1 is bounded from above, there exists a C > 0 such that an ≤ C for
all n ≥ 1. Let c be the smallest positive constant such that an ≤ c for all
n ≥ 1. We will show that lim an = c.
n→∞
Let > 0. Then there is an integer N such that (See Figure 8.1.3).
aN > c − .
Figure 8.1.3
Since the sequence {an }∞
n=1 is increasing, we find
an > c − for all n ≥ N. Thus,
|an − c| = |c − an | < for all n ≥ N. This shows that the sequence converges to c
The following is a list of useful sequences.
Theorem 8.1.5
(a) If r = 1 then lim rn = 1.
n→∞
(b) If r = −1 then the sequence {rn }∞
n=1 is divergent.
(c) If |r| > 1 then the sequence {rn }∞
n=1 is divergent.
n
(d) If |r| < 1 then lim r = 0.
n→∞
1
(e) For r > 0, lim r n = 1.
n→∞
Proof.
(a) Trivial.
(b) If r = −1 then the sequence {(−1)n }∞
n=1 alternates between the two
values −1 and 1 and so the sequence is divergent.
(c) |r| > 1 implies r > 1 or r < −1. Suppose first that r > 1. Let > 0. Let
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N be a positive integer greater than
r−1 .
Then for n ≥ N we have
rn =(1 + (r − 1))n
≥1 + n(r − 1)(by the binomial formula)
>1 + N (r − 1)
>1 + >.
This shows that for any given positive number we can find a term in the
n
sequence {rn }∞
n=1 which is greater than the number. This means that r →
∞ as n → ∞.
If r < −1 then rn = (−1)n (−r)n with −r > 1. Thus, as n becomes large, rn
alternates between large positive numbers and negative numbers with large
absolute value so that again the limit rn as n → ∞ does not exist.
1
with (r−1 )n → ∞ as n → ∞. (See (c)).
(d) If 0 < r < 1 then rn = (r−1
)n
Hence, rn → 0 as n → ∞.
If −1 < r < 0 then 0 < −r < 1. In this case, rn = (−1)n (−r)n → 0 as
n → ∞. If r = 0 then rn = 0 and limn→∞ rn = 0.
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(e) Let d = r n . Then ln d = lnnr → 0 as n → ∞. Hence, d = eln d → e0 = 1
as n → ∞.
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