MTH 231: CALCULUS III
Dr. Chao Huang
Department of Mathematics and Statistics
Chapter 1. In…nite Series
Section 1.1 In…nite Sequence
De…nition of Sequences: A (in…nite) sequence is a list of in…nite many
numbers in a pre-de…ned order. For instance,
1; 2; 3; 4; 5; 6; 7; 8; :::
1 1 1 1 1
1; ; ; ; ; ; :::
2 3 4 5 6
2 3 4 5 6
; ¡ ; ; ¡ ; ; :::
1 2 3 4 5
Note that since it is impossible to list in…nite many numbers, we may stop at
a certain point and omit the rest as long as a pattern is clear. For instance,
the above sequences could be written as
1; 2; 3; 4; :::;
n
; :::
(nth position)
1 1 1 1
; ; ; ; :::;
1 2 3 4
1
n
; :::
(nth position)
2 3 4 5
n+1
; ¡ ; ; ¡ ; :::(¡1)n+1
; :::
1 2 3 4
n
(nth position)
1
In general, a sequence can be expressed as
a1 ; a2 ; a3 ; a4 ; :::; an ; ::: or simply fan g1
n=1
a1 is called the …rst term,
a2 is called the second term ...
an is called the nth term
an+1 is the term right after an :
an¡1 is the term right before an :
Sometimes, for convenience, a sequence can be written as
fbn g1
n=4 : b4 ; b5 ; b6 ; :::
1
or fcn gn=¡3 : c¡3 ; c¡2 ; c¡1 ; c0 ; c1 ; :::
In many cases, a sequence an can be treated as a function de…ned for integer
numbers, an = f (n) :
Example 1.1.The three sequences introduced in the beginning,
1; 2; 3; 4; 5; 6; 7; 8; :::
1 1 1 1 1
1; ; ; ; ; ; :::
2 3 4 5 6
2 3 4 5 6
; ¡ ; ; ¡ ; ; :::
1 2 3 4 5
can be de…ned by, respectively,
an = f (n) = n; n = 1; 2; 3; ::: or fng1
½ n=1
¾
1
1 1
an = f (n) = ; n = 1; 2; 3; ::: or
n
n n=1
½
¾1
n+1 n + 1
n+1 n + 1
an = f (n) = (¡1)
; n = 1; 2; 3::: or (¡1)
:
n
n
n=1
Example 1.2. Some sequences de…ned by formulas:
½
¾1
n+1
1 2 3 4
1
fan gn=0 =
:
; ; ; ; :::
3n
30 3 32 33
n=0
n
n¼ o1
fan g1
=
cos
n=0
6 n=0
2
fan g1
n=3 =
p p p
©p
ª1
n ¡ 3 n=3 : 0; 1; 2; 3; :::
The last sequence has another equivalent expression:
©p ª1
fan g1
n n=0
n=0 =
Example 1.3. Some sequences may be de…ned by verbal descriptions:
an is the population of the world as of January 1st, year n
For instance, a2008 is the population of the world as of January 1st, 2008.
Example 1.4. Some sequences may be de…ned recursively:
(a) Fibonacci sequences: f0 = 0; f1 = 1; fn = fn¡1 + fn¡2 ; for n = 2; 3; :::
0; 1; 1; 2; 3; 5; 8; 13; 21; 34; :::
(b) Average sequence: a0 = 0; a1 = 1; an =
an¡1 + an¡2
; for n = 2; 3; :::
2
1 3 5 11 21
0; 1; ; ; ; ; ; :::
2 4 8 16 32
Graph of Sequence
The graph of a sequence fan g1
n=0 consists of a sequence of in…nite many
1
points f(n; an )gn=0 :
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In this example, y = 1 is the "asymptote" or limit of the sequence.
If a sequence is de…ned through a function
an = f (n) ;
then the relation between the graph of the function y = f (x) and the graph
of the sequence fan g1
n=0 is demonstrated in the following …gure:
solid line — graph of function y = f (x) ; dots — graph of sequence fan g
Limit of Sequence
Since sequences consist of in…nite many numbers, one needs to understand
their asymptotic behaviors, or "virtual end numbers".
De…nition 1.1. The limit of a sequence fan g1
n=1 , if existed, is de…ned
as a …nite number L such that an can be as close to L as one wants as long
as n is larger than a chosen number. More precisely, for any small number
" > 0 (for instance, " = 0:001); one can choose a very large number N (")
(this number becomes extremely large when " is very very small) satisfying
jan ¡ Lj < "
for any n > N (") :
This de…nition is consistent with the limit de…nition for functions. In particular, if an = f (n) is a sequence de…ned by a function y = f (x) ; then
lim an = lim f (x) = L:
n!1
x!1
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Example 1.5. Sequence limits. (a)
ln n
ln x
= lim
n!1 n
x!1 x
lim
L0 Hospital0 s rule
=
(ln x)0
1=x
=0
0 = lim
x!1 (x)
x!1 1
lim
(b)
2
2
5+ + 2
5n2 + 2n + 2
(5n2 + 2n + 2) =n2
n n =5
lim
= lim
= lim
3
n!1
n!1
n!1
n2 + 3
(n2 + 3) =n2
1+ 2
n
When we deal with limits of sequences, in general, we may apply
"leading term principle":
P (n)
leading term of P (n)
= lim
n!1 Q (n)
n!1 leading term of Q (n)
lim
For instance,
5n2 + 2n + 2
leading term of (5n2 + 2n + 2)
=
lim
n!1
n!1
n2 + 3
leading term of (n2 + 3)
5n2
= lim 2 = 5
n!1 n
lim
3n5 + 2n3 + 2n2 ¡ 5
3n5
3
=
lim
=
lim
=0
n!1
n!1 n6
n!1 n
n6 + 3n5 + 6
n4 + 3n3 ¡ 2n2 + 5
n4
lim
=
lim
= lim n2 = 1
n!1
n!1 n2
n!1
n2 + 3n ¡ 6
This leading term principle can also be applied "locally" at various levels.
For instance,
p
p
n4 + 3n3 ¡ 2n2 + 5
n4
lim
=
lim
= 1:
n!1
n!1 n2
n2 + 3n ¡ 6
lim
Example 1.6. Geometric sequence fr n g1
n=0 : For instance,
r = 2 : 1; 2; 22 ; 23 ; 24 ; ::: ¡! 1
1
1 1 1 1
r = : 1; ; 2 ; 3 ; 4 ; ::: ¡! 0:
2
2 2 2 2
5
In general,
n
lim r =
n!1
½
0;
1
if jrj < 1
if r > 1
Example 1.7. Alternating sequences.
an = (¡1)n
1
1 1 1
: ¡1; ; ¡ ; ; ::: ¡! 0
n
2 3 4
bn = (¡1)n : ¡1; 1; ¡1; 1; ¡1; 1; ::: It has no limit.
De…nition 1.2. A sequence fan g is called increasing if
an · an+1 :
If an can be represented by a function f (n) ; then sequence fan g is increasing
means the underline function f (x) is increasing. A sequence fan g is called
decreasing if
an ¸ an+1 :
A sequence fan g is called monotonic if it is either increasing or decreasing.
De…nition 1.3. A sequence fan g is called bounded above if there is a
…nite number M such that
an · M for all n:
The following sequence is bounded above:
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The following is not bounded above:
De…nition 1.4. A sequence fan g is called bounded below if there is
a …nite number m such that
m · an for all n:
A sequence fan g is called bounded if it is both bounded above and
bounded below.
Theorem. (a) Any bounded monotonic sequence is convergent.
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(b) Squeeze Theorem. If
bn · an · cn
lim bn = lim cn = L:
n!1
n!1
Then
lim an = L:
n!1
(c) Since
¡ jan j · an · jan j ;
lim an = 0 if and only if lim jan j = 0
n!1
n!1
Example 1.8. Consider the sequence:
a1 = 2; an+1 =
an + 6
:
2
We show by induction that
0 · an · 6 (bounded)
an · an+1
(increasing).
When n = 1; both statements above are true, since a2 = (2 + 6) =2 = 4.
Suppose they are true for n · k; i.e.,
0 · an · 6 for n · k
an · an+1
for n · k.
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We need to show they hold for n = k + 1; i.e.,
0 · ak+1 · 6
ak+1 · ak+2 :
(1)
(2)
To show the …rst state (1), we note that since (by assumption)
0 · ak · 6;
ak + 6
6+6
·
= 6:
2
2
The second statement (2) is veri…ed by
ak+1 =
ak+1 =
ak + 6
ak+1 + 6
·
· ak+2 :
2
2
According to Theorem, the sequence is convergent, i.e., there exists a number
L such that
lim an = L:
n!1
Now, if we apply limit to
an + 6
an + 6
=) lim an+1 = lim
n!1
n!1
2
2
lim an + 6
L+6
=) lim an+1 = n!1
=) L =
=) 2L = L + 6
n!1
2
2
So, lim an = L = 6:
an+1 =
n!1
Remark: Changing a …nite many items in a sequence doesn’t change
the limit. For instance, consider two sequences
(a) b1 ; b2 ; :::; b1000 ; a1001 ; a1002 ; :::an ; :::
(b) c1 ; c2 ; :::; c1000 ; a1001 ; a1002 ; :::an ; :::
Their …rst 1000 terms are di¤erent, but the rest terms are the same. But the
change in the …rst 100 terms does not change their asymptotic behaviors. In
fact, both sequences become identical when n > 1000: So these two sequences
either both convergent or both divergent. If convergent, they have the same
limit.
Lab Assignment (due at the end of the lab session): Lab #70
Homework:
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1. List …rst 5 terms of the sequence de…ned by
n
; n = 10; 11; 12; :::
2n + 1
5bn ¡ 2
(b) b0 = 2; bn+1 =
; n = 0; 1; 2; :::
2
(a) an = (¡1)n
2. Find a formula for the general term an of the following geometric sequence:
2 4
8
(a) 1; ¡ ; ; ¡ ; :::
3 9
27
(b) 3:14; 3:1415; 3:141515; 3:14151515; :::
an
(c) a0 = 3; an+1 = ; for n = 0; 1; 2; :::
3
3. Determine whether the sequence converges or diverges. If converges,
…nd the limit.
2n3 + 1
n3 + 2n2 + 3n + 5
n
(b) bn = (¡1)n
n+2
n
(c) cn = (¡1)n 2
n +2
2
(ln n)
(d) dn =
n
(a) an =
4. Find the …rst 40 terms of the sequence de…ned by
an+1
½ an
;
if an is an even number
2
=
3an + 1; if an is an odd number
and a1 = 11: Do the sam for a1 = 25: Then make a conjecture for this
type of series with any a1 .
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