LECTURE 27: SEQUENCES
MINGFENG ZHAO
March 16, 2015
Sequence
Definition 1. A sequence {an }∞
n=1 is an ordered list of numbers o the form
{a1 , a2 , a3 , · · · , }.
A sequence maybe be defined with an explicit formula of the form an = f (n) for n = 1, 2, 3, · · · . A sequence also may
be generated by a recurrence relation of the form an+1 = f (an ), for n = 1, 2, 3, · · · , where a1 is given.
Example 1. . Let an =
a1 = −
(−1)n n
for all n ≥ 1, then
n2 + 1
1
1
=− ,
12 + 1
2
a2 =
2
2
= ,
22 + 1
5
a3 = −
3
3
=− ,
32 + 1
10
and a10 =
Example 2. Let an+1 = 2an + 1 and a1 = 1, then
a2
=
2a1 + 1 = 2 · 1 + 1 = 3
a3
=
2a2 + 1 = 2 · 3 + 1 = 7
a4
=
2a3 + 1 = 2 · 7 + 1 = 15
Question 1. Can you find an explicit formula of an in Example 1?
In fact, since an+1 = 2an + 1, then
an+1 + 1 = 2an + 1 + 1 = 2an + 2 = 2(an + 1).
Let bn = an + 1, then bn+1 = 2bn and b1 = a1 + 1 = 2. So we get
bn =
bn
bn−1
·
bn−1
b2
···
· b1 = 2n−1 · 2 = 2n .
bn−2
b1
Therefore, we have
an = bn − 1 = 2n − 1,
1
for all n ≥ 1.
10
10
=
.
102 + 1
101
2
MINGFENG ZHAO
Limit of a sequence
Definition 2. If the terms of a sequence {an } approach a unique number L as n increase, that is, if an can be made
arbitrarily close to L by taking n sufficiently large, then we say that the sequence {an } converges to L, denoted by
lim an = L.
n→∞
If the terms of the sequence do not approach a single number as n increases, the sequence has no limit, and the
sequence diverges.
Example 3. Consider the sequence
(−1)n
n2 + 1
∞
, then
n=1
1
− ,
2
1
,
5
−
1
,
10
1
,··· .
17
The terms decrease in magnitude and approach 0 with alternating signs. So the limit appears to be 0.
Example 4. Consider the sequence {cos(nπ)}∞
n=1 , then
cos(π) = −1,
cos(3π) = −1,
cos(2π) = 1,
cos(4π) = 1, · · · .
The terms of the sequence alternate between 1 and -1, and never approaches a single value. So the sequence diverges.
Theorem 1. Suppose f (x) is a function such that f (n) = an for all positive integers n, if lim f (x) = L, then the limit
x→L
of the sequence {an } is also L.
Example 5. Find the limit of the sequence
Let f (x) =
3n3
an = 3
n +1
∞
.
n=1
3x3
, then f (n) = an for all n ≥ 1. Notice that
+1
x3
lim f (x) = lim
x→∞
x→∞
3x3
3
= lim
= 3.
3
x→∞
x +1
1 + x13
By Theorem 1, then
lim
n→∞
3n3
= 3.
n3 + 1
Theorem 2 (l’Hospital’s Rule). Let f (x) and g(x) be differential functions, then
I. Assume that lim f (x) = lim g(x) = 0 and lim
x→∞
x→∞
x→∞
lim
x→∞
f 0 (x)
= L, then
g 0 (x)
f (x)
= L.
g(x)
LECTURE 27: SEQUENCES
II. Assume that lim f (x) = ±∞, lim g(x) = ±∞ and lim
x→∞
x→∞
x→∞
f 0 (x)
= L, then
g 0 (x)
f (x)
= L.
g(x)
lim
x→∞
3
n+5
n
n ∞
.
Example 6. Find the limit of the sequence an =
n=1
x
x
x
x+5
5
x+5
, then f (n) = an for all n ≥ 1. Notice that for the function f (x) =
= 1+
,
Let f (x) =
5
x
x
when x → ∞, then lim f (x) has the form 1∞ . So we should consider ln f (x), then
x→∞
5
g(x) := ln f (x) = ln 1 +
x
x
ln 1 + x5
5
= x ln 1 +
=
.
1
x
x
Using l’Hospital’s Rule, we have
lim g(x)
x→∞
=
1
5
1+ x
lim
lim
x→∞
=
− x12
x→∞
=
· − x52
5
1+
5
x
5.
So we know that
lim f (x) = lim eg(x) = e5 .
x→∞
x→∞
By Theorem 1, then
lim
n→∞
n+5
n
n
= e5 .
o∞
n
1
.
Example 7. Find the limit of the sequence an = n n
n=1
1
1
Let f (x) = x x , then f (n) = an for all n ≥ 1. Notice that for the function f (x) = x x , when x → ∞, then lim f (x)
x→∞
0
has the form ∞ . So we should consider ln f (x), then
1
g(x) := ln f (x) = ln x x =
ln x
.
x
Using l’Hospital’s Rule, we have
lim g(x)
x→∞
=
lim
1
x
1
1
= lim
x→∞ x
=
x→∞
0.
4
MINGFENG ZHAO
So we know that
lim f (x) = lim eg(x) = e0 = 1.
x→∞
x→∞
By Theorem 1, then
1
lim n n = 1.
n→∞
Theorem 3 (Squeeze Theorem for Sequences). Let {an }, {bn } and {cn } be sequences such that an ≤ bn ≤ cn for all
integers n greater than some index N . If lim an = lim cn = L, then
n→∞
n→∞
lim bn = L.
n→∞
cos(n)
.
n2 + 1
Since −1 ≤ cos(n) ≤ 1, then
Example 8. Compute lim
n→∞
−
cos(n)
1
1
≤ 2
≤ 2
.
n2 + 1
n +1
n +1
Notice that
lim −
n→∞
n2
1
1
= 0 = lim
2
n→∞ n + 1
+1
By the squeeze theorem, Theorem 3, then
lim
n→∞
cos(n)
= 0.
n2 + 1
∞
Theorem 4. Assume that the sequences {an }∞
n=1 and {bn }n=1 have limits A and B, respectively, then
lim (an ± bn ) = A ± B,
n→∞
lim an bn = AB,
n→∞
lim can = cA,
n→∞
lim
n→∞
an
A
= ,
bn
B
where c is a real number
provided B 6= 0.
Example 9. By Example 5, Example 6 and Theorem 4, then
n 3n3
n+5
lim
−2
= 3 − 2 · e5 = 3 − 2et .
n→∞
n3 + 1
n
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca