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 a function from N into R. Notation. (1) The values of X : N ! R are denoted as X (n) or x n , where X is the sequence. (2) (x n : n 2 N) or simply (x n ) may denote a sequence — this is not the same as {x n : n 2 N}. (3) (x 1 ,x 2 ,...,x n ,... ). 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
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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}.
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.
|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.
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,
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 .