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CHAPTER 3
Sequences and Series
3.1. Sequences and Their Limits
Definition (3.1.1). A sequence of real numbers (or a sequence in R) is afunction from N into R.
Notation.
(1) The values of X : N ! R are denoted as X(n) or xn, where X is thesequence.
(2) (xn : n 2 N) or simply (xn) may denote a sequence — this is not the sameas {xn : n 2 N}.
(3) (x1, x2, . . . , xn, . . . ).
Example.
(1) (3n) = (3n : n 2 N) =
(3, 6, 9, . . . , 3n, . . . ).
(2) (1) = (1 : n 2 N) =(1, 1, 1, . . . , 1, . . . ).
(3)�(�2)n
�=�(�2)n : n 2 N
�=�
� 2, 4,�8, . . . , (�2)n, . . .�.
(4)⇣1
2+
1
2(�1)n
⌘=⇣1
2+
1
2(�1)n : n 2 N
⌘=
(0, 1, 0, 1, . . . , 0, 1, . . . ).
34
3.1. SEQUENCES AND THEIR LIMITS 35
(5)
✓⇣n
2
⌘12+1
2(�1)n◆
=
✓⇣n
2
⌘12+1
2(�1)n
: n 2 N◆
=
⇣1, 1, 1, 2, 1, 3, 1, 4, . . . , 1,
n
2, . . .
⌘.
Sequences may also be defined inductively or recursively.Example.
(1) x1 = 5, xn+1 = 2xn � 3 (n � 1) gives
(5, 7, 11, 19, 35, . . . ).
(2) Fibonacci sequence: x1 = x2 = 1, xn+1 = xn�1 + xn (n � 2) gives
(1, 1, 2, 3, 5, 8, 13, . . . ).
Definition (3.1.3). A sequence X = (xn) in R is said to converge to x 2 R,or have limit x, if
8 ✏ > 0 9 K(✏) 2 N 3��8 n � K(✏), |xn � x| < ✏.
We write this as lim X = x, lim(xn) = x, limn!1
xn = x, or
xn ! x as n !1.
A sequence that converges is called convergent, one that does not divergent.
Example. lim⇣1
n
⌘= 0.
Proof. Let ✏ > 0 be given. By the Archimedean property,
9 K(✏) 2 N 3�� 1
K(✏)< ✏. Then,
for n � K(✏),1
n 1
K(✏), and so
���1n� 0
��� =1
n 1
K(✏)< ✏.
Thus lim⇣1
n
⌘= 0 by definition. ⇤
36 3. SEQUENCES AND SERIES
Calculator Visualisation
limn!1
1
n= lim
x!1
1
x= 0 if 8 ✏ > 0,
with yMin = 0� ✏ and yMax = 0 + ✏,
you can find K(✏) 2 N 3�� if
xMin = K(✏) and xMax = 1E99,
the graph only enters the screen from the left and exits from the right.
Theorem (3.1.4 — Uniqueness of Limits). A sequence in R can have atmost one limit.
Proof. [The✏
2technique.]
Suppose lim(xn) = x0 and lim(xn) = x00. By Theorem 2.1.9,
it su�ces to show that |x0 � x00| < ✏ 8 ✏ > 0, for then
|x0 � x00| = 0 =) x0 = x00. Let ✏ > 0 be given.
Since lim(xn) = x0, 9 K 0 2 N 3�� 8 n � K 0, |xn � x0| <✏
2.
Since lim(xn) = x00, 9 K 00 2 N 3�� 8 n � K 00, |xn � x00| <✏
2.
Let K = max{K 0,K 00}. Then n � K =) n � K 0 and n � K 00 =)|x0 � x00| = |x0 � xn + xn � x00|| {z }
smuggling
|x0 � xn| + |xn � x00| <✏
2+
✏
2= ✏. ⇤
3.1. SEQUENCES AND THEIR LIMITS 37
Theorem (3.1.5). Let X = (xn) be a sequence in R, and let x 2 R. Thefollowing are equivalent:
(a) X converges to x.
(b) 8 ✏ > 0,9 K 2 N 3�� 8 n � K, |xn � x| < ✏.
(c) 8 ✏ > 0,9 K 2 N 3�� 8 n � K,x� ✏ < xn < x + ✏.
(d) 8 ✏-nbhd. V✏(x) of x,9 K 2 N 3�� 8 n � K,xn 2 V✏(x).Proof.
(a) () (b) by definition.
(b) () (c) () (d) since
|xn � x| < ✏ () �✏ < xn � x < ✏ () x� ✏ < xn < x + ✏ ()xn 2 V✏(x). ⇤
Technique
Given ✏ > 0. Produce or verify the existence of an integer K(✏) so thatn � K(✏) =) |xn � x| < ✏.
Sometimes |xn�x| < ✏ can be converted, with reversible steps, to an inequalityof the form n > f(✏). Take K(✏) as the first integer greater than f(✏) (by theArchimedean Property), K(✏) = [f(✏)] + 1, for example. Then
n � K(✏) =) n > f(✏) =) |xn � x| < ✏.Example.
(1) lim(c) = c, c 2 R, i,e., xn = c 8n 2 N.
Proof. Given ✏ > 0. [To show 9 K(✏) 2 N 3�� 8 n � K(✏), |c� c| < ✏.]
|c� c| = 0 < ✏ 8 n 2 N. Pick K(✏) = 1.
Then n � K(✏) =) |c� c| < ✏. ⇤
38 3. SEQUENCES AND SERIES
(2) lim⇣ 1p
n
⌘= 0. xn =
1pn
here.
Proof.
Given ✏ > 0.hTo show 9 K(✏) 2 N 3�� n � K(✏) =)
��� 1pn� 0
��� < ✏.i
Now
��� 1pn� 0
��� < ✏ () 1pn
< ✏ () 1
✏<p
n () 1
✏2< n.
Pick K(✏) =h 1
✏2
i+ 1. Then n � K(✏) =) n >
1
✏2=)
��� 1pn� 0
��� < ✏. ⇤
(3) lim⇣ c
np
⌘= 0, c 2 R, p > 0.
Proof. Case c = 0 was Example 1, so assume c 6= 0. Given ✏ > 0.��� c
np� 0
��� < ✏ () |c|np
< ✏ () |c|✏
< np ()⇣|c|
✏
⌘1/p< n.
Take K =
⇣|c|✏
⌘1/p�
+ 1.
Then n � K =) n >⇣|c|
✏
⌘1/p=)
��� c
np� 0
��� < ✏. ⇤
Note. Thus xn =1
3p
n, xn =
�5
n5/4, and xn =
1, 000, 000!
nall have limit 0.
3.1. SEQUENCES AND THEIR LIMITS 39
(4) lim⇣ 1
2n
⌘= 0.
Proof. Given ✏ > 0.��� 1
2n� 0
��� < ✏ () 1
2n< ✏ () 1
✏< 2n ()
ln1
✏< ln 2n () � ln ✏ < n ln 2 () � ln ✏
ln 2< n.
Take K = max
⇢1,h� ln ✏
ln 2
i+ 1
�.
Then n � K =) n >� ln ✏
ln 2=)
��� 1
2n� 0
��� < ✏. ⇤
(5) Let xn = 1 + (�1)n. X = (0, 2, 0, 2, . . . ).
lim(xn) does not exist.
Proof. [We use contradiction.]
Suppose lim(xn) = x. Then, 8 ✏ > 0,9 K 2 N 3�� 8 n � K, |xn � x| < ✏.
In particular, for ✏ = 1, 9 K 2 N 3�� 8 n � K, |xn � x| < 1.
But
(|0� x| < 1 for n odd
|2� x| < 1 for n even,
so 2 = |2� x + x| |2� x| + |x| |2� x| + |x� 0| < 1 + 1 = 2,
a contradiction.
Thus lim(xn) does not exist. ⇤
40 3. SEQUENCES AND SERIES
(6) Let xn =p
n. lim(xn) does not exist.
Proof. [We again use contradiction.]
Suppose lim(p
n) = x. Then, 8 ✏ > 0,9 K 2 N 3�� 8 n � K, |pn � x| < ✏or, equivalently, �✏ <
pn� x < ✏ or x� ✏ <
pn < x + ✏.
Then, for ✏ = 1, 8 n � K(1),p
n < x + 1 or n < (x + 1)2, contradicting theArchimedean Property.
Thus lim(p
n) does not exist. ⇤
Homework
Pages 61-62 #5b,5d (Do not use Theorem 3.1.10 with these — work from thedefinition)
Extra Problem: Prove�(�1)n
�diverges. (Hint: This is a translation of Exam-
ple 5 — watch your inequalities, though.)
Note. Sometimes it is awkward or impossible to solve |xn � x| < ✏ for n.In such cases, it may be possible to establish an inequality of the form
|xn � x| C|an|where C > 0 and lim an = 0.
3.1. SEQUENCES AND THEIR LIMITS 41
Theorem (3.1.10). Let (an) and (xn) be sequences in R, lim(an) = 0,and x 2 R. If for some C > 0 and some m 2 N we have
|xn � x| C|an| 8 n � m,
then lim(xn) = x.
Proof. Let ✏ > 0 be given. Since lim(an) = 0,
9 Ka
⇣ ✏
C
⌘2 N 3�� 8 n � Ka
⇣ ✏
C
⌘, |an � 0| <
✏
C.
Let Kx(✏) = max
⇢m,Ka
⇣ ✏
C
⌘�. Then
8 n � Kx(✏), |xn � x| |{z}n � m
C|an| <|{z}n � Ka
⇣ ✏
C
⌘C · ✏
C= ✏.
Thus lim(xn) = x. ⇤Example.
(7) lim⇣1 + (�1)n
n
⌘= 0.
Proof.hX =
⇣0, 1, 0,
1
2, 0,
1
3, . . . , 0,
1
n, . . .
⌘i.
���1 + (�1)n
n� 0
��� =���1 + (�1)n
n
��� 1 + 1
n=
2
n�! 0
by Example 3. The result follows from Theorem 3.1.10. ⇤
42 3. SEQUENCES AND SERIES
(8) lim⇣ n + 1
3n + 2
⌘=
1
3.
Proof.hX =
⇣2
5,3
8,
4
11,
5
14,
6
17,
7
20, . . . ,
n + 1
3n + 2, . . .
⌘i.
��� n + 1
3n + 2� 1
3
��� =���3n + 3� 3n� 2
3(3n + 2)
��� =1
3(3n + 2) 1
3(3n)=
19
n�! 0
by Example 3. The result follows from Theorem 3.1.10. ⇤
(9) lim⇣ n + 1
n3 +p
n
⌘= 0.
Proof.hX =
⇣1,
3
8 +p
2,
4
27 +p
3,
5
66,
6
125 +p
5, . . . ,
n + 1
n3 +p
n, . . .
⌘i.
��� n + 1
n3 +p
n� 0
��� =n + 1
n3 +p
n
n + 1
n3
n + n
n3=
2n
n3=
2
n2�! 0
by Example 3. The result follows from Theorem 3.1.10. ⇤
3.1. SEQUENCES AND THEIR LIMITS 43
(10) lim⇣ n3 + 3
2n3 � n
⌘=
1
2.
Proof.hX =
⇣4,
11
14,10
17,
67
124,122
245, . . . ,
n3 + 3
2n3 � n, . . .
⌘i.
��� n3 + 3
2n3 � n� 1
2
��� =���2n3 + 6� 2n3 + n
2(2n3 � n)
��� =n + 6
2(2n3 � n)|{z}
n � 6n + n
2(2n3 � n3)=
2n
2n3=
1
n2�! 0
by Example 3. The result follows from Theorem 3.1.10. ⇤
“Ultimate Behavior”
Definition (3.1.8). If X = (x1, x2, . . . , xn, . . . ) is a sequence in R and ifm 2 N, the m-tail of X is the sequence
Xm = (xm+n : n 2 N) = (xm+1, xm+2, . . . , xm+n, . . . ).
Example. The 4-tail of⇣1,
1
2,1
3, . . . ,
1
n, . . .
⌘is
X4 =⇣1
5,1
6,1
7, . . . ,
1
4 + n, . . .
⌘.
Theorem (3.1.9). Let X = (xn : n 2 N) be a sequence and let m 2 N.Then the m-tail Xm = (xm+n : n 2 N) converges () X converges. Inthis case,
lim Xm = lim X.
Proof. Read in text — it is just a translation argument. ⇤
Homework
Pages 61-62 #6a, 6c, 10, 11
Extra: Prove lim(p
n + 1�pn) = 0.
Hint for #10: Look at Theorem 3.1.5(c) and pick the right ✏.
44 3. SEQUENCES AND SERIES
3.2. Limit Theorems
Definition (3.2.1). A sequence (xn) is bounded if
9 M > 0 3�� |xn| M 8 n 2 N.
Theorem (3.2.2). If (xn) converges, then (xn) is bounded.
Proof. Suppose lim(xn) = x and ✏ = 1.
Then 9 K(1) 2 N 3�� 8 n � K(1), |xn � x| < 1.
Then, for n � K(1),
|xn| = |xn � x + x| |xn � x| + |x| < 1 + |x|.Let M = sup
�|x1|, |x2|, . . . , |xK(1)�1|, 1 + |x|
.
Then |xn| M 8 n 2 N. ⇤
Example.�(�1)n
�is bounded since |(�1)n| 1 8 n 2 N, but does not
converge. Thus, bounded 6=) convergent.
Example. (2n) diverges.
Proof. If (2n) converged, it would be bounded.
Thus 9 M > 0 3�� |2n| = 2n M 8 n 2 N.
Then n = log2 2n log2 M 8 n 2 N,
contradicting the Archimedean Property. ⇤
3.2. LIMIT THEOREMS 45
Theorem (3.2.3). Suppose lim(xn) = x and lim(yn) = y.
(a) lim(xn + yn) = lim(xn) + lim(yn) = x + y.
Proof. [The✏
2technique.]
Since lim(xn) = x, 9 K1 2 N 3�� 8 n � K1, |xn � x| <✏
2.
Since lim(yn) = y, 9 K2 2 N 3�� 8 n � K2, |yn � y| <✏
2.
Let K = max{K1,K2}. Then n � K =) n � K1 and n � K2 =)|(xn + yn)� (x + y)| = |(xn � x) + (yn � y)|
|xn � x| + |yn � y| <✏
2+
✏
2= ✏. ⇤
(b) lim(xn � yn) = lim(xn)� lim(yn) = x� y.
Proof. Similar to the above. ⇤
46 3. SEQUENCES AND SERIES
(c) lim(xnyn) = lim(xn) lim(yn) = xy.
Proof. Let ✏ > 0 be given. Note
|xnyy � xy| = (by smuggling)
|(xnyn � xny) + (xny � xy)| |(xnyn � xny)| + |(xny � xy)|
|xn||yn � y| + |y||xn � x|.[We are now able to gain control over all of the variable parts.]
Since lim(xn) = x:
(1) 9 M1 > 0 3�� |xn| M1 8 n 2 N by Theorem 3.2.2.
Let M = max�M1, |y|
.
(2) 9 K1 2 N 3�� 8n � K1, |xn � x| <✏
2M.
Since lim(yn) = y, 9 K2 2 N 3�� 8n � K2, |yn � y| <✏
2M.
Let K = max�K1,K2
. Then, 8n � K,
|xnyy � xy| |xn||yn � y| + |y||xn � x| M · ✏
2M+ M · ✏
2M= ✏.
⇤
(d) lim(cxn) = c lim(xn) = cx for c 2 R.
Proof. This is a special case of (c). ⇤
3.2. LIMIT THEOREMS 47
(e) If lim(zn) = z, zn 6= 0 8 n 2 N, and z 6= 0, then lim⇣ 1
zn
⌘=
1
z.
Proof. Let ✏ > 0 be given. Note��� 1
zn� 1
z
��� =���z � zn
znz
��� =1
|znz|· |zn � z|.
hWe need to find a bound for
1
|zn|in the first factor.
i
Let ↵ =1
2|z| > 0. Since lim(zn) = z:
(1) 9 K1 2 N 3�� 8n � K1, |zn � z| < ↵. Then
�↵ < �|zn � z| |{z}Cor.2.2.4(a)Th.2.2.2(c)
|zn|� |z| =)
1
2|z| = |z|� ↵ |zn|| {z }
|zn| is bounded away from 0
=) 1
|zn| 2
|z|.
(2) 9 K2 2 N 3�� 8n � K2, |zn � z| <1
2✏|z|2.
Let K = max�K1,K2
. Then, 8n � K,��� 1
zn� 1
z
��� =1
|znz|· |zn � z| <
2
|z|2⇣1
2✏|z|2
⌘= ✏.
⇤
(f) If lim(zn) = z, zn 6= 0 8 n 2 N, and z 6= 0, then lim⇣xn
zn
⌘=
lim(xn)
lim(zn)=
x
z.
Proof. This follows directly from parts (c) and (e) above. ⇤
48 3. SEQUENCES AND SERIES
Example. Find lim⇣3n2 � 2
n2 + n
⌘.
Proof.
lim⇣3n2 � 2
n2 + n
⌘= lim
✓3� 2
n2
1 + 1n
◆=
lim�3� 2
n2
�lim
�1 + 1
n
� =
lim(3)� lim�
2n2
�lim(1) + lim
�1n
� =3� 0
1� 0= 3.
⇤
Theorem (3.2.4). If lim(xn) = x and xn � 0 8 n 2 N, then x � 0.
Proof. [Use contradiction by picking an appropriate ✏.]
Suppose x < 0 =) �x > 0. Since lim(xn) = x,
for ✏ = �x, 9 K 2 N 3�� 8 n � K, |xn � x| < �x or
�(�x) < xn � x < �x or x + x < xn < �x + x = 0.
Thus, for n = K, xK < 0, contradicting our hypotheses.
Thus x � 0. ⇤
Theorem (3.2.5). If (xn) and (yn) are convergent sequences and if xn yn
8 n 2 N, then lim(xn) lim(yn).
Proof. Let zn = yn � xn. Then zn � 0 8 n 2 N,
so 0 lim(zn) = lim(yn)� lim(xn) =) lim(xn) lim(yn). ⇤
Theorem (3.2.6). If (xn) is convergent and a xn b 8 n 2 N, thena lim(xn) b.
Proof. This follows from Theorem 3.2.5 by comparing (a) and (b) with(xn). ⇤
3.2. LIMIT THEOREMS 49
Theorem (3.2.7 — Squeeze Theorem). Suppose xn yn zn 8n 2 Nand lim(xn) = lim(zn). Then (yn) converges and
lim(xn) = lim(yn) = lim(zn).
Proof. Let w = lim(xn) = lim(zn). Given ✏ > 0.
9 K1 2 N 3�� 8 n � K1, �✏ < xn � w < ✏, and also
9 K2 2 N 3�� 8 n � K2, �✏ < zn � w < ✏.
Let K = max�K1,K2
. Then for n � K,
�✏ <|{z}n � K1
xn � w yn � w zn � w <|{z}n � K2
✏ =) |yn � w| < ✏.
Thus lim(yn) = w. ⇤
Note. The hypotheses of Theorem 3.2.4 thru Theorem 3.2.7 can be weak-ened to apply to tails of the sequences rather than to the sequences themselves.
Example.
(1) Find lim⇣cos n
n
⌘.
Solution. �1 cos n 1 =) �1
n cos n
n 1
n.
Since lim⇣� 1
n
⌘= lim
⇣1
n
⌘= 0,
lim⇣cos n
n
⌘= 0 by the Squeeze Theorem. ⇤
50 3. SEQUENCES AND SERIES
(2) Find lim�n1/n
�.
Solution. [This one is tricky.]
For n > 1, n1/n > 1, so xn = n1/n = 1 + tn, where tn = n1/n � 1 > 0. Then,
from the Binomial theorem,
n = (1 + tn)n = 1 + ntn +
n(n� 1)
2t2n + positive terms,
son(n� 1)
2t2n < n =) t2n <
2
n� 1=) tn <
p2p
n� 1.
Thus
1 < xn = 1 + tn < 1 +
p2p
n� 1=) 1 < xn < 1 +
p2p
n� 1.
Since lim(1) = lim⇣1 +
p2p
n� 1
⌘= 1,
lim(xn) = lim�n1/n
�= 1 by the Squeeze Theorem. ⇤
(3) Find lim⇣
np
n2�.
Solution.
lim� np
n2�
= lim�
np
n · np
n�
= lim�
np
n�· lim
�np
n�
= 1 · 1 = 1.
⇤
Theorem (3.2.9). Suppose lim(xn) = x. Then lim(|xn|) = |x|.Proof. We know
��|xn|� |x|�� |xn � x|. Thus, given ✏ > 0,
if 9 K 2 N 3�� 8 n � K, |xn � x| < ✏, we also get��|xn|� |x|
�� < ✏. ⇤
3.2. LIMIT THEOREMS 51
Theorem (3.2.10). Suppose lim(xn) = x and xn � 0 8 n 2 N. Thenlim(
pxn) =
px.
Proof. [Using the conjugate].
First, x � 0 by Theorem 3.2.4. Let ✏ > 0 be given.
Case x = 0 9 K 2 N 3�� 8 n � K, |xn � 0| < ✏2 ()0 xn < ✏2 () 0 pxn < ✏ () |pxn � 0| < ✏.
Case x > 0 Thenp
x > 0. 9 K 2 N 3�� 8 n � K, |xn � x| <p
x✏ =)
|pxn �p
x| =
����(p
xn �p
x) · (p
xn +p
x)p
xn +p
x
���� =
|xn � x|pxn +
px |xn � x|p
x<
px✏px
= ✏.
⇤
Homework
Pages 69-70 #1d, 5b, 6bd (both find and prove)
Extra: If lim⇣xn
n
⌘= x 6= 0, then (xn) is not bounded.
⇣Hint: Prove (xn)
bounded =) lim⇣xn
n
⌘= 0
⌘.
52 3. SEQUENCES AND SERIES
3.3. Monotone Sequences
Definition (3.3.1). Let X = (xn) be a sequence.
X is increasing if x1 x2 · · · xn xn+1 · · · .X is decreasing if x1 � x2 � · · · � xn � xn+1 � · · · .X is monotone if it is either increasing or decreasing.
Example.
(1) (1, 1, 2, 3, 5, 8, . . . ) is increasing.
(2) For 0 < b < 1, (b, b2, b3, . . . ) is decreasing.
(3) (2, 0, 2, 0, . . . ) is not monotone.
(4) (4, 2, 1, 3, 3, 5, 5, 7, 7, . . . ) is ultimately increasing.
Theorem (3.3.2 — Monotone Convergence Theorem (MCT)). A mono-tone sequence converges () it is bounded. Further:
(a) if X = (xn) is a bounded, increasing sequence, then lim(xn) = sup{xn}.(b) if X = (xn) is a bounded, decreasing sequence, then lim(xn) = inf{xn}.
Proof. (=)) Follows from Theorem 3.2.2.
((=) (a) Since (xn) is bounded,
9 M > 0 3�� |xn| M 8 n 2 N =) xn M 8 n 2 N.
Thus, by Completeness, x? = sup{xn} exists. Let ✏ > 0 be given.
By Property S, 9 K 2 N 3�� x? � ✏ < xK (xK = s✏).
Since (xn) is increasing, xK xn 8 n � K. Thus, 8 n � K,
x? � ✏| {z } < xK xn|{z} x? < x? + ✏| {z } =) |xn � x?| < ✏.
Thus lim(xn) = x?.
(b) Similar to (a), except uses Property I. ⇤
3.3. MONOTONE SEQUENCES 53
Example. Determine whether lim(xn) exists and, if so, its value wherex1 = 1 and xn+1 =
p1 + xn for n � 1.
Solution.
x2 =p
1 + 1 =p
2, x3 =
q1 +
p2, x4 =
r1 +
q1 +
p2 , . . .
(a) [Show monotone increasing.]
x1 < x2 since 1 <p
2. Assume xn xn+1.
Then xn+1 =p
1 + xn p
1 + xn+1 = xn+2,
so by induction xn xn+1 8 n 2 N.
Thus (xn) is increasing.
(b) [Show (xn) is bounded above by 2 using induction.]
x1 = 1 < 2. Suppose xn 2. Then
xn+1 =p
1 + xn p
1 + 2 =p
3 <p
4 = 2.
Thus, by induction, xn 2 8 n 2 N,
and so 2 is an upper bound of (xn).
(c) Thus lim(xn) = x for some x 2 R by the MCT.
Since (xn+1) is a tail of (xn), lim(xn+1) = x also. Then
x = lim(xn+1) = lim(p
1 + xn) =p
lim(1 + xn) =plim(1) + lim(xn) =
p1 + x =)
x2 = 1 + x =) x2 � x� 1 = 0 =) x =1 ±
p5
2.
Since1�
p5
2< 0, we conclude x = lim(xn) =
1 +p
5
2. ⇤
54 3. SEQUENCES AND SERIES
Note.
(1) An increasing sequence is bounded below by its first term. Thus if x? =sup{xn : n 2 N},
M = max�|x1|, |x?|
is a bound for the sequence.
(2) A decreasing sequence is bounded above by its first term.
Homework
Page 77 # 1, 2
Hint for # 2: (a) Show xn � xn+1 � 0 8 n 2 N. Thus (xn) is decreasing.
(b) Show (xn) is bounded below. Then (xn) is bounded by M = max�|x1|, |l.b.|
.
(c) Find and solve an equation to get x = lim(xn).
Example.
(2) Establish convergence or divergence of (xn) where
xn = 1 +1
1!+
1
2!+ · · · +
1
n!.
Solution. xn+1 = xn +1
(n + 1)!> xn, so (xn) is increasing.
Noting that1
(n + 1)!<
1
2n, we have
xn < 1 +⇣1 +
1
2+
1
22+
1
23+ · · · +
1
2n�1
⌘=
1 +1� (1
2)n
1� 12
= 1 + 2�⇣1
2
⌘n�1< 3,
so (xn) is bounded above and so (xn) converges by the MCT.
Although we now know the limit exists, we do not have a technique for findingthe exact limit. ⇤
3.3. MONOTONE SEQUENCES 55
(3)
Let An = the sum of the semicircular areas.
Let Ln = the sum of the semicircumferences.
It appears limn!1
An = 0 and limn!1
Ln = 1. Well,
An = n · ⇡
2·⇣ 1
2n
⌘2=
⇡
8· 1
n! 0 as n !1,
but
Ln = n · ⇡ ·⇣ 1
2n
⌘=
⇡
28 n 2 N,
so limn!1
Ln =⇡
2.
56 3. SEQUENCES AND SERIES
Problem (Page 77 # 10). Establish convergence or divergence of (yn) where
yn =1
n + 1| {z }largestterm
+1
n + 2+ · · · +
1
2n|{z}smallest
term
8 n 2 N.
Solution. It might seem obvious that lim(yn) = 0, but incorrect.
Note that
yn �1
2n+ · · · +
1
2n| {z }n terms
= n · 1
2n=
1
2
and
yn 1
n + 1+ · · · +
1
n + 1| {z }n terms
= n · 1
n + 1=
n
n + 1< 1,
so (yn) is bounded by 1.
Now
yn+1 =1
n + 2+
1
n + 3+ · · · +
1
2n+
1
2n + 1+
1
2n + 2,
so
yn+1 � yn =1
2n + 1+
1
2n + 2� 1
n + 1=
1
2n + 1+
1
2n + 2� 2
2n + 2=
1
2n + 1� 1
2n + 2=
1
(2n + 1)(2n + 2)> 0,
so (yn) is increasing. Thus (yn) converges by the MCT and
1
2 lim(yn) 1.
Can we find lim(yn)?
3.3. MONOTONE SEQUENCES 57
Note
yn =nX
k=1
1
n + k=
nXk=1
1
1 + kn
· 1
n.
[What is this latter sum?]
yn is a right-hand Riemann sum for
f(x) =1
1 + xfor 0 x 1.
Thus
lim(yn) =
Z 1
0
1
1 + xdx = ln|1 + x|
���10
= ln 2� ln 1 = ln 2.
⇤
58 3. SEQUENCES AND SERIES
3.4. Subsequences and the Bolzano-Weierstrass Theorem
Definition (3.4.1). Let X = (xn) be a sequence and let
n1 < n2 < · · · < nk < · · ·be a strictly increasing sequence of natural numbers. Then the sequence
X 0 = (xnk) = (xn1, xn2, . . . , xnk
, . . . )
is a subsequence of X.
Example. Let X =⇣ 1
2n
⌘=⇣1
2,1
4,1
6, . . . ,
1
2n, . . .
⌘.
Some subsequences:
(1) (xnk) =
⇣ 1
4k
⌘=⇣1
4,1
8,
1
12, . . . ,
1
4k, . . .
⌘.
(2) (xnk) =
⇣ 1
4k � 2
⌘=⇣1
2,1
6,
1
10, . . . ,
1
4k � 2, . . .
⌘.
(3) (xnk) =
⇣ 1
4k2
⌘=⇣1
4,
1
16,
1
36, . . . ,
1
4k2, . . .
⌘.
(4) (xnk) =
⇣ 1
(2k)!
⌘=⇣ 1
2!,
1
4!,
1
6!, . . . ,
1
(2k)!, . . .
⌘=⇣1
2,
1
24,
1
720, . . . ,
1
(2k)!, . . .
⌘.
(5) X itself
(6) Any tail of X
In general, to form a subsequence of X, just pick out any infinite selection ofterms of X going from left to right.
3.4. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 59
Theorem (3.4.2). If X = (xn) converges to x, so does any subsequence(xnk
).
Proof. Let ✏ > 0 be given. Since (xn) converges to x,
9 K 2 N 3�� 8 n � K, |xn � x| < ✏. Since
n1 < n2 < · · · < nk < · · ·is an increasing sequence in N, nk � k 8 k 2 N.
Let K 0 = nK . Then, 8 nk � K 0 = nK, nk � K =) |xnk� x| < ✏.
Thus (xnk) converges to x. ⇤
Example. For c > 1, find lim(c1n) if it exists.
Solution.
(a) xn = c1n > 1 8 n 2 N, so (xn) is bounded below.
(b) xn � xn+1 = c1n � c
1n+1 = c
1n+1
�c
1n(n+1) � 1
�> 0 8 n 2 N,
so (xn) is decreasing.
(c) Thus lim(xn) = x exists.
[Using a subsequence to find x.]
Now x2n = c12n =
�c
1n�1
2 =�xn
�12 , so
x = lim(x2n) = lim��
xn
�12�
= x12 =)
x2 = x =) x2 � x = 0 =) x(x� 1) = 0 =) x = 0 or x = 1.
Since xn > 1 8 n 2 N, lim(xn) = 1. ⇤
60 3. SEQUENCES AND SERIES
Theorem (3.4.7 — Monotone Subsequence Theorem). If X = (xn) is asequence in R, then there is a subsequence of X that is monotone.
Proof. We will call xm a peak if n � m =) xn xm (i.e, if no term tothe right of xm is greater than xm).
Case 1 : X has infinitely many peaks.
Order the peaks by increasing subscripts. Then
xm1 � xm2 � · · · � xmk� · · · ,
so(xm1, xm2, . . . , xmk
, . . . )
is a decreasing subsequence.
Case 2 : X has finitely many (maybe 0) peaks.
Let xm1, xm2, . . . , xmr denote these peaks.
Let s1 = mr + 1 (the first index past the last peak) or s1 = 1 if there are nopeaks.
Since xs1 is not a peak, 9 s2 > s1 3�� xs1 < xs2.
Since xs2 is not a peak, 9 s3 > s2 3�� xs2 < xs3.
Continuing, we get an increasing subsequence. ⇤
Theorem (3.4.8 — Bolzaono-Weierstrass Theorem). A bounded sequenceof real numbers has a convergent subsequence.
Proof. If X = (xn) is bounded, by the Monotone Subsequence Theorem ithas a monotone subsequence X 0 which is also bounded. Then X 0 is convergentby the MCT. ⇤
3.4. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 61
Theorem (3.4.4). Let X = (xn) be a sequence. The following are equiv-alent:
(a) (xn) does not converge to x 2 R.
(b) 9 ✏0 > 0 3�� 8 k 2 N, 9 nk 2 N 3�� nk � k and |xnk� x| � ✏0.
(c) 9 ✏0 > 0 and a subsequence X 0 = (xnk) of X 3�� |xnk
�x| � ✏0 8 k 2 N.Proof.
[(a) =) (b)] This is the negative of the definition of convergence.
[(b) =) (c)] Take the ✏0 from (b).
Let n1 2 N 3�� |xn1 � x| � ✏0.
Let n2 2 N 3�� n2 > n1 and |xn2 � x| � ✏0.
Let n3 2 N 3�� n3 > n2 and |xn3 � x| � ✏0.
Continuing, we generate the subsequence.
[(c) =) (a)] Suppose X = (xn) has a subsequence X 0 = (xnk) satisfying (c).
If xn ! x, so would (xnk) ! x. Then 9 K 2 N 3�� 8k � K, |xnk
� x| < ✏0.
But this contradicts (c). ⇤
62 3. SEQUENCES AND SERIES
Example.⇣
cosn⇡
4
⌘does not converge to
p2
2.
Proof.⇣
cosn⇡
4
⌘=⇣p2
2, 0,�
p2
2,�1,�
p2
2, 0,
p2
2, 1, . . .
⌘.
Let ✏0 =
p2
4. 8 k 2 N, let nk = 8k + 3.
Then (xnk) =
⇣cos
(8k + 3)⇡
4
⌘=⇣�p
2
2
⌘.
Then 8 k 2 N,���xnk
�p
2
2
��� =����
p2
2�p
2
2
��� =p
2 �p
2
4= ✏0.
Thus⇣
cosn⇡
4
⌘does not converge to
p2
2. ⇤
Theorem (3.4.5 — Divergence Criterion). If a sequence X = (xn) haseither of the following properties, then X is divergent.
(a) X has two convergent subsequences X 0 = (xnk) and X 00 = (xrk
) whoselimits are not equal.
(b) X is unbounded.
Homework
Pages 84-85 # 4b, 9 (Hint: Use Theorem 3.4.4), 11 (Hint: What is the onlypossible limit?), 14 (extra credit)
3.4. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 63
Theorem (3.4.9). Let X = (xn) be a bounded sequence such that everyconvergent subsequence converges to x. Then lim(xn) = x.
Proof. Let M be a bound for X. Suppose xn 6! x. By Theorem 3.4.4,
9 ✏0 > 0 and a subsequence X 0 = (xnk) 3�� |xnk
� x| � ✏0 8 k 2 N.
Now M is also a bound for X 0 = (xnk),
so it has a convergent subsequence X 00 = (xnkr) with lim(xnkr
) = x.
Then 9 K 2 N 3�� 8 r � K, |xnkr� x| < ✏0, a contradiction. ⇤
Example. We cannot drop the bounded hypothesis:⇣1,
1
2, 3,
1
4, 5,
1
6, . . .
⌘.
64 3. SEQUENCES AND SERIES
3.5. The Cauchy Criterion
Example. Suppose lim(xm � xn) = 0? Does lim(xn) necessarily exist?
NO! xn =p
n is a counterexample.
Definition (3.5.1). A sequence X = (xn) is a Cauchy sequence if
8 ✏ > 0 9 H(✏) 2 N 3�� 8 n,m � H(✏) with n,m 2 N, |xn � xm| < ✏
.
Lemma (3.5.3). If X = (xn) converges, then X is Cauchy.
Proof. [Another✏
2argument.]
Suppose lim(xn) = x. Given ✏ > 0, 9 K 2 N 3�� 8 n � K, |xn � x| <✏
2.
Let H = K. Then, for m,n � H = K,
|xn � xm| = |(xn � x) + (x � xm)| |xn � x| + |xm � x| <✏
2+
✏
2= ✏
Thus (xn) is Cauchy. ⇤
Note. (xn) is not Cauchy if
9 ✏0 > 0 3�� 8H 2 N, 9 n,m � H 3�� |xn � xm| � ✏0
.
Example. (xn) =p
n is not Cauchy.
Proof. Let ✏0 = 1 and H 2 N be given.
Let m = H, sop
xm =p
m =p
H.
Since�p
n�
is unbounded,
9 n 2 N 3�� |p
n�p
H| � 1
where n � H.
Thus |pxn �p
xm| � 1. ⇤
3.5. THE CAUCHY CRITERION 65
Lemma (3.5.4). Cauchy sequences are bounded.
Proof. Let X = (xn) be Cauchy and ✏ = 1
9 H 2 N 3�� 8 n � H, |xn � xH| < 1.
Then ��|xn|� |xH|�� |xn � xH| < 1 =)
�1 < |xn|� |xH| < 1 =) |xn| < |xH| + 1.
Let M = max�|x1|, |x2|, . . . , |xH�1|, |xH| + 1
.
Then |xn| M 8n 2 N. ⇤
Theorem (3.5.5 — Cauchy Convergence Criterion). A sequence is con-vergent () it is Cauchy.
Proof. [Yet another✏
2argument.]
(=)) Lemma 3.5.3
((=) Let X = (xn) be Cauchy. Then X is bounded,
so by B-W, X has a convergent subsequence, say X 0 = (xnk) ! x?.
[To show lim(xn) = x?.] Let ✏ > 0 be given.
Since (xn) is Cauchy, 9 H 2 N 3�� 8 n,m � H, |xn � xm| <✏
2.
Since lim(xnk) = x?, 9 K 2 N 3�� K � H and |xK � x?| <
✏
2.
But |xn � xK| <✏
2also. Then, for n � H,
|xn�x?| = |(xn�xK) + (xK �x?)| |(xn�xK)|+ |(xK �x?)| <✏
2+
✏
2= ✏.
Thus lim(xn) = x?. ⇤
66 3. SEQUENCES AND SERIES
Example. (xn) =⇣1
n
⌘is Cauchy.
Proof.
Given ✏ > 0. WLOG (without loss of generality), suppose n � m.
Then ���1n� 1
m
��� =1
m� 1
n<
1
m< ✏ (=|{z}
ifm >
1
✏.
Take H =h1
✏
i+ 1. Then
n � m � H =) n � m >1
✏=)
���1n� 1
m
��� < ✏.
⇤
3.5. THE CAUCHY CRITERION 67
Problem (Page 91 # 2b). Show⇣1 +
1
2!+
1
3!+ · · · +
1
n!
⌘is Cauchy.
Proof.hIn this proof we use the facts that 2n�1 n! (Example 1.2.4(e)) and that
1 + r + r2 + · · · + rn =1� rn+1
1� r.i
Given ✏ > 0. WLOG, suppose n � m.
|xn � xm| =���⇣1 +
1
2!+
1
3!+ · · · +
1
n!
⌘�⇣1 +
1
2!+
1
3!+ · · · +
1
m!
⌘��� =��� 1
(m + 1)!+
1
(m + 2)!+ · · · +
1
n!
��� 1
2m+
1
2m+1+ · · · +
1
2n�1
1
2m
⇣1 +
1
2+ · · · 1
2n�m�1
⌘=
1
2m·1�
�12
�n�m
1� 12
=2n�m � 1
2n�1
2n�m
2n�1=
1
2m�1< ✏ (=
1
✏< 2m�1 (= log2
1
✏< m� 1 (= 1� log2 ✏ < m.
Choose H = maxn
1,h1� log2 ✏
i+ 1
o. Then
n � m � H =) m > 1� log2 ✏ =) |xn � xm| < ✏.
⇤
Definition (3.5.7). A sequence X = (xn) is contractive if
9 0 < C < 1 3�� |xn+2 � xn+1| C|xn+1 � xn| 8 n 2 N.
C is the constant of the contractive sequence.
68 3. SEQUENCES AND SERIES
Theorem (3.5.8). Every contractive sequence is convergent.
Proof. [We prove the sequence to be Cauchy, and thus convergent.]
Let X = (xn) be a contractive sequence. 8 n 2 N,
|xn+2 � xn+1| C|xn+1 � xn| C2|xn � xn�1| · · · Cn|x2 � x1|.Then, WLOG for m > n,
|xm � xn| |xm � xm�1| + |xm�1 � xm�2| + · · · + |xn+1 � xn|| {z }smuggling + triangle inequality
�Cm�2 + Cm�3 + · · · + Cn�1
�|x2 � x1| =
Cn�1�Cm�n�1 + Cm�n�2 + · · · + 1
�|x2 � x1| =
Cn�1⇣1� Cm�n
1� C
⌘|x2 � x1|
Cn�1⇣ 1
1� C
⌘|x2 � x1|! 0 as n !1
since lim(Cn) = 0. Thus (xn) is Cauchy, and so convergent. ⇤
3.5. THE CAUCHY CRITERION 69
Example. x1 = 1, x2 = 2, xn =1
2(xn�2 + xn�1) for n � 3.
(xn) =⇣1, 2,
3
2,7
4,13
8,27
16, . . .
⌘.
(a) (xn) is contractive. Thus (xn) converges.
Proof.
|xn+2 � xn+1| =���12(xn + xn+1)� xn+1
��� =���12xn �
1
2xn+1
��� =1
2|xn+1 � xn|.
⇤
(b) Note that
|xn+1 � xn| =1
2n�1|x2 � x1| =
1
2n�1and x2n+1 � x2n�1 > 0 (by induction).
(c) [To find lim(x2n+1) = lim(xn).]
x2n+1 � x2n�1 =1
2(x2n�1 + x2n)� x2n�1 =
1
2x2n �
1
2x2n�1 =
1
2|x2n � x2n�1| =
1
2· 1
22n�2=
1
22n�1.
Thus
x2n+1 = x2n�1 +1
22n�1= x2n�3 +
1
22n�3+
1
22n�1= · · · =
1 +1
2+
1
23+
1
25+ · · · +
1
22n�1= 1 +
1
2
h1 +
1
22+
1
24+ · · · +
1
22n�2
i=
1 +1
2
h1 +
1
22+
1
24+ · · · +
1
22n�2
i= 1 +
1
2
h1 +
1
4+⇣1
4
⌘2+ · · · +
⇣1
4
⌘n�1i=
1 +1
2·1�
�14
�n
1� 14
= 1 +1
2·4� 1
4n�1
4� 1=
1 +4� 1
4n�1
6! 1 +
2
3=
5
3as n !1.
Thus lim(xn) = lim(x2n+1) =5
3.
70 3. SEQUENCES AND SERIES
3.6. Properly Divergent Sequences
Definition. Let (xn) be a sequence.
(a) We say (xn) tends to +1 and write lim(xn) = +1 if
8 ↵ 2 R 9 K(↵) 2 N 3�� 8 n � K(↵), xn > ↵.
(b) We say (xn) tends to �1 and write lim(xn) = �1 if
8 � 2 R 9 K(�) 2 N 3�� 8 n � K(�), xn < �.
We say (xn) is properly divergent in either case.
Example. For C > 1, lim(Cn) = +1Proof. Let ↵ 2 R be given. [How to express C > 1.]
C = 1 + b where b > 0. By the Archimedean Property,
9 K(↵) 2 N 3�� K(↵) >↵
b. Then 8 n � K(↵),
Cn = (1 + b)n �|{z}Bernoulli
1 + nb > 1 + ↵ > ↵.
Thus lim(Cn) = +1. ⇤
Homework
Page 91 # 2a, 3b, 7, 9
3.7. INTRODUCTION TO INFINITE SERIES 145
3.7. Introduction to Infinite Series
Definition (3.7.1). If X = (xn) is a sequence in R, then the infinite series(or just series) generated by X is the sequence S = (sk) defined by
s1 = x1
s2 = s1 + x2 (= x1 + x2)...
sk = sk�1 + xk (= x1 + x2 + · · · + xk)...
The xn are the terms of the series and the sk are the partial sums of the series.If lim S exists, we say the series is convergent and call this limit the sum orvalue of the series. If this limit does not exist, we say this series S is divergent.
Notation. X(xn) or
Xxn or
1Xn=1
xn
We can also use 1Xn=0
xn or1X
n=5
xn
If the first term of the series is xN , then the first partial sum is sN .
146 3. SEQUENCES AND SERIES
Example.
(1)1X
n=0
rn = 1 + r + r2 + · · · + rn + · · · (geometric series)
sn = 1 + r + r2 + · · · + rn
rsn = r + r2 + · · · + rn + rn+1
sn(1� r) = sn � rsn = 1� rn+1
For r 6= 1,
sn =1� rn+1
1� rThen 1X
n=0
rn = limn!1
1� rn+1
1� r=
1
1� rif |r| < 1,
and1X
n=0
rn diverges if |r| � 1.
(2)1X
n=1
1
n(n + 1)=
1
1 · 2+
1
2 · 3+
1
3 · 4+ · · · .
Since1
k(k + 1)=
1
k� 1
k + 1,
sn =⇣1� 1
2
⌘+⇣1
2� 1
3
⌘+⇣1
3� 1
4
⌘+ · · · +
⇣1
n� 1
n + 1
⌘=)
sn = 1� 1
n + 1=)
1Xn=1
1
n(n + 1)= lim
n!1
⇣1� 1
n + 1
⌘= 1.
3.7. INTRODUCTION TO INFINITE SERIES 147
(3)1X
n=0
(�1)n = 1� 1 + 1� 1 + · · · .
S = (sn) = (1, 0, 1, 0, . . . ) diverges =)1X
n=0
(�1)n diverges.
Theorem (3.7.3 —nth Term Test). IfP
xn converges, lim(xn) = 0.
Proof.P
xn converges =) s = lim(sn) exists =)s = lim(sn�1) =)
lim(xn) = lim(sn � sn�1) = lim(sn)� lim(sn�1) = s� s = 0.
⇤Example.
(4) Geometric series with |r| � 1 diverges since (rn) diverges.
(5) For1X
n=1
1pn
= 1 +1p2
+1p3
+ · · · +1pn
+ · · · ,
lim(xn) = lim⇣ 1p
n
⌘= 0.
But
sn = 1 +1p2
+1p3
+ · · · +1pn
� 1pn
+1pn
+1pn
+ · · · +1pn| {z }
n terms
= n · 1pn
=p
n
Thus lim(sn) � lim(p
n) !1, so1X
x=1
1pn
diverges.
Note. This implies the converse of Theorem 3.7.3 is not true.
148 3. SEQUENCES AND SERIES
(6) Consider1X
n=1
cos n.
Assume lim(cos n) = 0 =) lim(cos2 n) = 0 =)lim(sin2 n) = lim(1� cos2 n) = lim(1)� lim(cos2 n) = 1� 0 = 1.
Then lim(sin2 2n) = 1 as a subsequence.
Now sin 2n = 2 sin n cos n =) sin2 2n = 4 sin2 n cos2 n =)lim(sin2 2n) = 4 lim(sin2 n) lim(cos2 n) = 4 · 1 · 0 = 0 6= 1,
a contradiction.
Thus lim(cos n) 6= 0 =)1X
n=1
cos n diverges.
Theorem (3.7.4 — Cauchy Criterion for Series).Pxn converges () 8 ✏ > 0 9 M 2 N 3�� if m > n � M , then
|sm � sn| = |xn+1 + xn+2 + · · · + xm| < ✏.
Theorem (3.7.5). Let (xn) be a sequence of nonnegative numbers. ThenPxn converges () S = (sk) is bounded. In this case,
nXi=1
xn = lim(sk) = sup{sk : k 2 N}.
Proof. Since xn � 0 8 n 2 N, (sk) is monotone increasing.
By the MCT, S = (sk) converges () it is bounded, in which case
lim(sk) = sup{sk}.⇤
3.7. INTRODUCTION TO INFINITE SERIES 149
Example.
(7) The harmonic series1X
n=1
1
ndiverges. The proof is similar to that of Example
5.
(8) The p-series1X
n=1
1
npconverges for p > 1.
Proof. Since (sk) is monotone, we need only to show (sk) is bounded. Butit su�ces to show that some subsequence is bounded.
Let k1 = 21 � 1 = 1, so sk1 = 1.
Let k2 = 22 � 1 = 3. Then, since 2p < 3p,
sk2 =1
1p+⇣ 1
2p+
1
3p
⌘< 1 +
2
2p= 1 +
1
2p�1.
Let k3 = 23 � 1 = 7 =)
sk3 = sk2 +⇣ 1
4p+
1
5p+
1
6p+
1
7p
�
< sk2 +4
4p< 1 +
1
2p�1+
1
4p�1.
Continuing inductively, if kj = 2j � 1,
0 < skj = 1 +1
2p�1+
1
4p�1+ · · · +
1
(2j�1)p�1
= 1 +1
2p�1+
1
(2p�1)2+ · · · +
1
(2p�1)j�1
<1
1� 12p�1
.
Thus (skj) is bounded and so1X
n=1
1
npconverges for p > 1. ⇤
150 3. SEQUENCES AND SERIES
(9) The p-series1X
n=1
1
npdiverges for p 1.
For p 0,1
np= n�p with �p � 0 =) lim
⇣ 1
np
⌘6= 0.
For 0 < p 1, np < n =) 1
n 1
np.
Since the partial sumes of the harmonic series are not bounded, neither are the
partial sums here. Thus1X
n=1
1
npdiverges for p 1.
Theorem (9.3.2 — Alternating Series Test). Let Z = (zn) be a decreasingsequence of strictly positive numbers with lim(zn) = 0. Then the alternating
series1X
n=1
(�1)n+1zn converges.
Proof. s2n = (z1 � z2) + (z3 � z4) + · · · + (z2n�1 � z2n),
and since zk � zk+1 > 0, (s2n) is increasing.
Since s2n = z1 � (z2 � z3)� (z4 � z5)� · · ·� (z2n�2 � z2n�1)� z2n,
s2n < z1 8 n 2 N =) (s2n) converges by the MCT.
Suppose lim(s2n) = s. Then, given ✏ > 0, 9 K 2 N 3�� for n � K,
|s2n � s| <✏
2and |z2n+1| <
✏
2.
Then for n � K,
|s2n+1 � s| = |s2n + z2n+1 � s| |s2n � s| + |z2n+1| <✏
2+
✏
2= ✏.
Thus, for n large enough, each partial sum is within ✏ of s, so lim(sn) = s and1X
n=1
(�1)n+1zn = s. ⇤
3.7. INTRODUCTION TO INFINITE SERIES 151
Example.
(10) The alternating harmonic series1X
n=1
(�1)n+1 1
nconverges.
Note. If1X
n=1
(�1)n+1zn = s, |s� sn| zn+1.
For1X
n=1
(�1)n+1 1
n,
���s�10X
n=1
(�1)n+1 1
n
��� =���s� 1627
2520
��� <1
11.
Theorem (3.7.7 — Comparison Test). Let (xn) and (yn) be sequenceswhere 0 xn yn for n � K 2 N. Then
(a)P
yn converges =)P
xn converges.
(b)P
xn diverges =)P
yn diverges.
Proof.
(a) SupposeP
yn converges and ✏ > 0 is given.
9 M 2 N 3�� m > n � M =) yn+1 + · · · + ym < ✏.
Let M 0 = max{K,M}. Then m > M 0 =)0 xn+1 + · · · + xm yn+1 + · · · + ym < ✏,
soP
xn converges.
(b) Contrapositive of (a). ⇤
152 3. SEQUENCES AND SERIES
Example.
(11)1X
n=1
1
n2 + nconverges since
1
n2 + n<
1
n28 n 2 N and
1Xn=1
1
n2converges (p = 2).
(12)1X
n=1
1
n2 � n + 1. Seems to be like
1Xn=1
1
n2.
But1
n2 � n + 1>
1
n28n 2 N.
We can show 0 <1
n2 � n + 1 2
n28n 2 N , but this is not obvious.
Theorem (3.7.8 — Limit Comparison Test). Suppose (xn) and (yn) are
strictly positive sequences and suppose r = lim⇣xn
yn
⌘exists.
(a) If r 6= 0,P
xn converges ()P
yn converges.
(b) If r = 0 andP
yn converges, thenP
xn converges.Proof.
(a) Since r = lim⇣xn
yn
⌘6= 0, 9 K 2 N 3��
1
2r xn
yn 2r for n � K =)
⇣1
2r⌘yn xn (2r)yn for n � K.
The result follows by applying comparison twice.
(b) If r = 0, 9 K 2 N 3�� 0 < xn yn for n � K.
The result follows by comparison. ⇤
3.7. INTRODUCTION TO INFINITE SERIES 153
Example.
(12 continued)1X
n=1
1
n2 � n + 1.
Since
r = lim
✓ 1n2�n+1
1n2
◆= lim
⇣ n2
n2 � n + 1
⌘= lim
✓1
1� 1n + 1
n2
◆= 1,
1Xn=1
1
n2 � n + 1converges by limit comparison since
1Xn=1
1
n2converges.
(13)1X
n=1
1p3n + 1
.
We compare with1X
n=1
1pn
=1X
n=1
1
n1/2, which diverges as a p-series with p 1.
r = lim
✓ 1p3n+11pn
◆= lim
⇣ n1/2
(3n + 1)1/2
⌘= lim
✓h n
3n + 1
i1/2◆
=
lim
⇣ n
3n + 1
⌘�1/2
=⇣1
3
⌘1/2=
1p36= 0.
Thus1X
n=1
1p3n + 1
diverges by limit comparison.
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