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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= ✏. ⇤


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| < ✏. ⇤


(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.


(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. ⇤


(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.


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. ⇤


(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


(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



(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


“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 ✏.


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. ⇤


(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). ⇤


(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. ⇤


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). ⇤


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. ⇤


(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|

�� < ✏. ⇤


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

⌘.


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. ⇤


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)

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.


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)?


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.

⇤


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. ⇤


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. ⇤


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). ⇤


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)


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, . . .

⌘.


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?. ⇤


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

�� < ✏.

⇤


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.


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. ⇤


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.


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 .


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)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.


(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}.⇤


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. ⇤


(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. ⇤


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). ⇤


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. ⇤


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.

sequences and series - christian brothers universityfacstaff.cbu.edu/wschrein/media/m414...

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