Ism et chapter_8

(c)

2

1.8

1.6

Chapter 8 1.4

1.2

1

Inﬁnite Series
0.8

0 2 4 6 8 10 12
x

n 1
7. (a) lim = lim 1 =1
n→∞ n + 1 n→∞ 1 +
8.1 Sequences of Real n

Numbers (b) As n gets large, n/(n + 1) gets close to 1.

3 5 7 9 11 (c)
1. 1, , , , ,
4 9 16 25 36
0.9

3 1 3 3 1 3
2. , , , , ,
5 2 7 8 3 10 0.8

2 1 1 1
3. 4, 2, , , , 0.7
3 6 30 180

1 2 3 4 5 6 0.6
4. − , , − , , − ,
2 3 4 5 6 7
0.5

1 0 2 4 6 8 10 12

5. (a) lim 3 = 0 x
n→∞ n

(b) As n gets large, n3 gets large, so 1/n3 goes 2n + 1 1
to 0. 8. (a) lim = lim 2 + =2
n→∞ n n→∞ n
(c)
(b) As n gets large, (2n + 1)/n gets close to
1 2.

0.8 (c)

0.6

0.4

0.2

0
0 2 4 6 8 10 12
x

2
6. (a) lim √ = 0
n→∞ n
√ 2
(b) As n gets large, n gets large, so √ goes
n 1
to 0. 9. (a) In Q.(5), an =
n3

443

444 CHAPTER 8. INFINITE SERIES

1.0
3.0

2.5
0.75

2.0

0.5
1.5

1.0
0.25

0.5

0.0
0.0
2 4 6 8 10
5 10 15 20

As n gets larger and larger, the terms As n gets larger and larger, the terms
1
of the sequence an = n3 approach to 0. of the sequence an = 2n+1 approach 2.
n
Hence, an converges to 0. Hence, the sequence an = 2n+1 converges
n
to 2.
2
(b) In Q.(6), an = √ 10.
n
1.0

0.9

0.8 0.5

0.7
0.0
0.6
0 10 20 30 40

0.5
−0.5

0.4

0.3 −1.0

10 20 30 40 50
As n gets larger, the terms of the sequence
As n gets larger and larger, the terms n nπ nπ
2 an = sin + cos approach to 1
of the sequence an = √n approach to 0. n+1 2 2
2 or -1. When n is of the form 4m or 4m + 1, the
Hence, the sequence an = √n converges
terms of the sequence an approach 1. When
to 0.
n is of the form 4m + 2 or 4m + 3, the terms
n of the sequence an approach -1. Therefore, an
(c) In Q.(7), an = does not convergent.
n+1
1
3n2 + 1 3+ n2 3
11. lim 2−1
= lim 1 =
n→∞ 2n n→∞ 2 − 2
0.9 n2
1
5n3 − 1 5− 3
0.8
12. lim = lim n =5
n→∞ 2n3 + 1 n→∞ 1 2
0.7 2+ 3
n
1
0.6 n2 + 1 n+ n
13. lim = lim =∞
n→∞ n + 1 n→∞ 1 + 1
n
0.5
5 10 15 20 25 30 1 1
n2 + 1 + 3
As n gets larger and larger, the terms 14. lim 3 = lim n n = 0
n→∞ n + 1 n→∞ 1
n
of the sequence an = n+1 approach 1. 1+ 3
n
n
Hence, the sequence an = n+1 converges 2
to 1. n+2 1+
15. lim (−1)n = lim (−1)n n
1
n→∞ 3n − 1 n→∞ 3− n
2n + 1 1
(d) In Q.(8), an = = ± , the limit does not exist; diverges
n 3

8.1. SEQUENCES OF REAL NUMBERS 445

n
16. Diverges. Even terms converge to 1 and odd = lim √
terms converge to −1. n→∞ n2 + n + n
1 1
= lim =
n+2
n n+2 n→∞ 1 + 1/n + 1 2
17. lim (−1) 2 = lim 2
n→∞ n +4 n→∞ n + 4
1
+ 22 27. lim [ln (2n + 1) − ln n]
= lim n n = 0 n→∞
n→∞ 1 + 4 2n + 1
n2
(−1)2 (n + 2) = lim ln
Hence lim =0 n→∞ n
n→∞ n2 + 4 1
= lim ln 2 + = ln 2
18. Diverges. Even terms are 1 and odd terms are
n→∞ n
−1. nπ
28. When n is even number, cos = 1 and
x 1 2
19. lim x = lim x by l’Hopital’s Rule nπ 2n − 1 2n − 1
x→∞ e x→∞ e lim cos = lim = 2.
1 n n→∞ 2 n+2 n→∞ n + 2
and lim x = 0, so by Theorem 1.2 lim n = nπ
x→∞ e n→∞ e When n is odd number, cos = 0 and
2
0 nπ 2n − 1
lim cos = 0.
1 cos n 1 n→∞ 2 n+2
20. Note that − n ≤ n ≤ n and Therefore, the limit does not exist.
e e e
1 cos n
lim = 0, so lim =0 29. By L’Hopital Rule,
n→∞ en n→∞ en
by the Squeeze Theorem. n3 + 1 3n2 6n
lim n
= lim n = lim n
1 2 n→∞ e n→∞ e n→∞ e
en + 2 n + e2n
6
21. lim = lim e 1 =0 = lim n = 0.
n→∞ e2n − 1 n→∞ 1 − 2n
e n→∞ e

22. For n ≥ 1, en + 1 < 2en , so 30. By L’Hopital Rule,
n
3n 1 3n 1 3 ln n 1
> · n = · . lim √ = lim n
en + 1 2 e 2 e n→∞ n + 1 n→∞
n √1
3 1 3 2 n+1
Since > 1, lim = ∞; √
e n→∞ 2 e 2 n+1
3n = lim
so lim n =∞ n→∞ n
n→∞ e + 1
1 1
= lim 2 + = 0.
x2x x x n→∞ n n2
23. lim = lim x = lim x
x→∞ 3x x→∞ 3 x→∞ 3
−1 cos n 1
2x 2 31. −1 ≤ cos n ≤ 1 ⇒ ≤ ≤ 2
1 n2 n2 n
= lim = 0, by l’Hopital’s Rule, −1 1
x→∞ 3
x
3 for all n, and lim 2 = lim 2 = 0
n→∞ n n→∞ n
ln
2 2 so by the Squeeze Theorem,
3
x cos n
since lim = ∞. lim =0
n→∞ n2
x→∞ 2
n
n2 1 cos πn 1 1
Hence lim n = 0, by Theorem 1.2. 32. Note that − 2 ≤ ≤ 2 and lim 2 =
n→∞ 3 n n2 n n→∞ n
cos πn
n! 1 2 n 1 n n 0, so lim = 0 by the Squeeze Theorem.
24. Since n
= · ··· ≥ · = and n→∞ n2
2 2 2 2 2 2 4
n n! 1 1 1
lim = ∞, we have lim n = ∞. 33. 0 ≤ |an | = n ≤ and lim =0
n→∞ 4 n→∞ 2 ne n n→∞ n
1
so by the Squeeze Theorem & Corollary 1.1,
1 sin n
25. lim n sin = lim =1 (−1)n
n→∞ n n→∞ n 1 lim =0
n→∞ nen

26. lim n2 + n − n 1
n→∞ √ √ ln x
n2 + n − n n2 + n + n 34. lim 2 = lim x = lim 1 = 0,
= lim √ x→∞ x x→∞ 2x x→∞ 2x2
n→∞ n2 + n + n by l’Hopital’s Rule.


ln n ln n ln n is not a lower bound for S, there exist some
Since (−1)n ≤ 2 , we have lim = ∞
n2 n n→∞ n2 aN , of for which aN < G + ε. Since {an }n=1
0, by the Squeeze Theorem and Corollary 1.1. is decreasing sequence, G < an < G + ε for
an+1 n+4 n+2 all n ≥ N . That is, G − ε ≤ an ≤ G + ε
∞
35. = · for all n ≥ N. Hence {an }n=1 converges to G.
an n+3 n+3
Therefore bounded and decreasing sequence is
n2 + 6n + 8
= 2 < 1 for all n, so convergent.
n + 6n + 9 √
an+1 < an for all n, so the sequence is decreas- 48. First we note that a2 = 3 + 2 3 > a1 . As-
ing. sume that ak > ak−1 , then
n n−1 √ √
36. Since an+1 − an = − ak+1 = 3 + 2 ak > 3 + 2 ak−1 = ak
n+2 n+1 Then we note that a1 < 3 and if we assume
2
= > 0, the sequence is increas- that ak−1 < 3 then
(n + 1)(n + 2) √ √
ing. ak = 3 + 2 ak−1 < 3 + 2 · 3 = 3
We have shown that the sequence is increas-
an+1 en+1 n e·n
37. = · n = > 1 for all n, ing with an upper bound, so it converges. The
an n+1 e n+1 limit of this sequence should be the solution to
∞
so an+1 > an for all n, so {an }n=1 is increasing. the equation
√
3n+1 x = 3 + 2x, x2 − 2x − 3 = 0
(x + 1)(x − 3) = 0, x = −1, x = 3 Since
an+1 (n + 3)! 3
38. Since = = ≤ 1, the se- the limit of the sequence can’t be negative, so
an 3n n+3 lim an = 3.
(n + 2)! n→∞
quence is decreasing. 49. (a) a1000 ≈ 7.374312390,
3n2 − 2 3n2 − 2 e2 ≈ 7.389056099
39. |an | = 2+1
= 2 b1000 ≈ .135064522,
n n +1
3n2 3n2 e−2 ≈ .135335283.
< 2 < 2 =3 m
n +1 n (b) Let n = , m = nr.
r
m/r
6n − 1 6n + 18 r n r2
40. |an | = < =6 lim 1 + = lim 1 +
n+3 n+3 n→∞ n m→∞ m
r
m/r 2
sin(n2 ) 1 1 1
41. |an | = ≤ ≤ for n > 1. = lim 1+ = er
n+1 n+1 2 m→∞ m/r2
as m/r2 → ∞
42. |an | = e1/n ≤ e1 = e
n−4
43. an = −(−2) 50. (a) For the ﬁrst 8 terms we get:
1
a1 = 1,
44. an = 2n−1 a2 = 2.5,
2n−1 a3 = 2.05,
45. an = n2 a4 = 2.000609756,
(−1)n+1 n a5 = 2.000000093,
46. an = (n+1)2 a6 = 2.000000000,
∞ a7 = 2.000000000,
47. Suppose that {an }n=1 is decreasing and
a8 = 2.000000000
bounded sequence. Then for some M > 0,
1 4
|an | < M for all n. Let S be the set con- The equation L = L+ has two
taining all of the terms of the sequence, S = 2 L
solutions, L = 2 and L = −2. Since the
{a1 , a2 , a3 , ...} . By Completeness Axiom, S
terms of the sequence are positive, we dis-
must have a greatest lower bound say G. That
card the negative solution. Thus the limit
is, G is the greatest number for which an > G
must be L = 2.
, for all n . Therefore, for any number ε >
0,G + ε > G and so,G + ε is not a lower bound, (b) Assuming the limit exists, and letting L =
since G is greatest lower bound. Since G + ε lim an+1 = lim an , we have
n→∞ n→∞

8.1. SEQUENCES OF REAL NUMBERS 447

1 c converges for p < −1 and p ≥ 1.
L= L+ ⇒ 2L2 = L2 + c
2 L √ 1
⇒ L2 = c ⇒ L = c. (b) p > 0 ⇒ lim =0
n→∞ np
1
51. Use induction to show that an+1 > an . First √ p = 0 ⇒ lim p = 1
n→∞ n
note that a2 = 2, and since a2 = 2 + 2,
1 2 1
it follows that a2 > a2 , so a2 > a1 . Thus, p < 0 ⇒ lim p does not exist
2 1 n→∞ n
the statement is true for n = 1. Now assume Therefore, the sequence an = 1/np
that the statement is true for n = k (that is, converges for p ≥ 0.
ak+1 > ak ), and show that the statement is
true for n = k + 1. First note that 1
√ 53. an = (1 + 2 + 3 + · · · + n)
ak+2 = 2 + ak+1 n2
√ 1 n(n + 1) n+1 1 1
a2 = 2 + ak+1 = 2 = = 1+
k+2
2 √ n 2 2n 2 n
ak+2 − 2 = ak+1 1 1 1
2 lim an = lim 1 + =
a2 − 2 = ak+1
k+2 n→∞ 2 n→∞ n 2
The previous equation is also valid if we replace Thus, the sequence an converges to 1/2.
2
k by k − 1: a2 − 2 = ak
k+1
Note that
n
1
Since ak+1 > ak , it follows that 1k
2 xdx = lim
ak+1 > a2 − 2 and therefore
k+1 0 n→∞
k=1
nn
2 2 Therefore the sequence an converges to
a2 − 2 > a2 − 2
k+2 k+1 1
a2 − 2 > a2 − 2
k+2 k+1 xdx.
a2 > a2 , ak+2 > ak+1
k+2 k+1 0
Thus, by induction, an is increasing. Now we’ll n
prove that an < 2 by induction. First note 1
54. lim an = lim
that a1 < 2, and assume that ak < 2. Then n→∞ n→∞ n+k
k=1
√ n
ak+1 = 2 + ak 1 n
√ 2 = lim
a2 − 2 = ak , a2 − 2 = ak
k+1 k+1
n→∞ nn+k
2 √ k=1
a2 − 2 < 2, a2 − 2 < 2
k+1
n
1 1
√ 2 k+1
a2 < 2 + 2, ak+1 < 4 = lim
k+1 n→∞ n 1 + k/n
ak+1 < 2 Thus, by induction, an < 2. k=1
1
Since an is increasing and bounded above by 1 1
= dx = ln |1 + x||0
2, an converges. To estimate the limit, we’ll
√ 0 1+x
approximate the solution of x =
√ 2 + x: = ln 2 − ln 1 = ln 2
x2 = 2 + x, (x2 − 2)2 = x Therefore the sequence converges.
x4 − 4x2 + 4 = x, 0 = x4 − 4x2 − x + 4
0 = (x − 1)(x3 + x2 − 3x − 4)
√ 55. (a) Begin by joining the centers of circles C1
Since an > 2, it follows that x = 1. and C2 with a line segment. The length
Therefore, 0 = x3 + x2 − 3x − 4. Using a CAS, of this line segment is the sum of the radii
the solution is x ≈ 1.8312. of the two circles, which is r1 + r2 . Thus,
1 the squared of the length of the line seg-
52. (a) p > 1 ⇒ lim =0 ment is (r1 +r2 )2 . Now, the coordinates of
n→∞ pn
1 the centers of the circles are (c1 , r1 ) and
p = 1 ⇒ lim n = 1 (c2 , r2 ). Using the formula for the dis-
n→∞ p
1 tance between two points, the square of
0 < p < 1 ⇒ lim n does not exist the length of the line segment joining the
n→∞ p
1 two centers is (c2 −c1 )2 +(r2 −r1 )2 . Equat-
− 1 < p < 0 ⇒ lim n does not exist ing the two expressions, we get
n→∞ p
1 (c2 − c1 )2 + (r2 − r1 )2 = (r1 + r2 )2
p = −1 ⇒ lim n does not exist Expanding and simplifying this relation-
n→∞ p
1 ship, we get
p < −1 ⇒ lim n = 0
n→∞ p (c2 − c1 )2 = (r1 + r2 )2 − (r2 − r1 )2
2 2 2 2
Therefore, the sequence an = 1/pn = r1 + 2r1 r2 + r2 − (r2 − 2r1 r2 + r1 )


= 4r1 r2 57. Given that s2n = 2 − 4 − sn 2 and
√
|c2 − c1 | = 2 r1 r2 s6 = 1. Therefore,
(b) The same reasoning applied to the other
two pairs of centers yields analogous re- S12 = 2− 4 − S6 2
sults. Without going through the motions √
again, you can simply take the results = 2− 4 − 12 = 2− 3
above and ﬁrst replace all subscripts “2”
by “3” to get the results for circles C1 S24 = 2− 4 − S12 2
and C3 . Then take these new results and
√
replace all subscripts “1” by “2” to get = 2− 4− 2− 3
the results for circles C2 and C3 . The
results are √
(c3 − c1 )2 + (r3 − r1 )2 = (r1 + r3 )2 = 2− 2+ 3
√
|c3 − c1 | = 2 r1 r3
(c3 − c2 )2 + (r3 − r2 )2 = (r2 + r3 )2 S48 = 2− 4 − S24 2
√
|c3 − c2 | = 2 r2 r3
√
= 2− 4− 2− 2+ 3
(c) Finally,
|c1 − c2 | = |c1 − c3 | + |c3 − c2 |
√ √ √
2 r1 r2 = 2 r1 r3 + 2 r2 r3 √
√ √ √ √ = 2− 2+ 2+ 3 ≈ 0.130806
r1 r2 = √ 3 ( r1 + r2 )
r
√ r1 r2 As S6 = 1, the length of each side of the regu-
r3 = √ √
r1 + r2 lar hexagon inscribed in the circle is 1 and the
(d) Given that r1 = r2 = 1, from Part(c), we corresponding angle subtended at the centre
π
have √ of the circle is , which implies the radius of
√ r1 r2 1 3
r3 = √ √ = the circle is 1. As n increases the length of the
r1 + r2
√ 2 arc is approximately equal to the length of the
√ r2 r3 1/2 1 2π
r4 = √ √ = = side of the regular n-gon. Therefore, Sn ≈
r2 + r3
√ 1 + 1/2 3 n
π
√ r3 r4 1/6 1 . Thus, S48 ≈ .
r5 = √ √ = = 24
r3 + r4 1/2 + 1/3 5
1 1 1
The pattern is √ = √ +√
rn rn−1 rn−2
Hence if Fn is the n-th Fibonacci number,
2
then rn = 1/Fn .

56. (a) The distance between the two points
(0, c) and (x0 , y0 ), where y0 = x2 , is r,
0

x2 + (x2 − c)2 = r
0 0
x2 + x4 − 2cx2 + c2 = r2
0 0 0
y0 + (1 − 2c)y0 + (c2 − r2 ) = 0
2 n! 1·2·3·4···n
58. 0 < n
=
We want the solution y0 to the above n n·n·n·n···n
equation to be unique, so that 1 2·3·4···n 1 1
= < (1) =
(1 − 2c)2 − 4(c2 − r2 ) = 0 n n·n·n···n n n
1 − 4c + 4c2 − 4c2 + 4r2 = 0 n! 1
1 Hence, 0 < n < .
1 − 4c + 4r2 = 0, c = + r2 n n
4 Therefore,
n! 1
(b) Following Part(a), the pattern is lim 0 < lim n < lim
n→∞ n→∞ n n→∞ n
2 1 1 2
rn + = rn + rn−1 + + rn−1 n!
4 4 0 < lim n < 0
2 2 n→∞ n
rn − rn−1 = rn + rn−1
rn − rn−1 = 1 n!
lim =0
r1 = 1, r2 = 2, r3 = 3, . . . rn = n n→∞ nn

8.2. INFINITE SERIES 449

2 0.5 61. In the 3rd month, only the adult rabbits have
3 0.5436 newborns, so a3 = 2+1 = 3. In the 4th month,
4 0.5731 only the 2 pairs of adult rabbits from a2 can
5 0.5949 have newborns, so a4 = 3 + 2 = 5. In general,
6 0.6119 an = an−1 + an−2
7 0.6258
8 0.6374 62. Let an and an+1 denote the sizes of the shorter
9 0.6473 and longer sides of the nth rectangle. For ex-
10 0.6559 ample, a1 = 1 and a2 = 2 are the sides of the
first rectangle, a2 = 2 and a3 = 3 are the sides
20 0.7065
of the second rectangle, and so on. When a new
30 0.7316
rectangle an ×an+1 is formed, the longer side of
40 0.7476
the previous one, an , becomes the shorter side
50 0.7590
of the new one, and the longer one of the new
60 0.7678
one, an+1 , is the sum of the shorter one and the
70 0.7748 longer one of the previous one. If we express
80 0.7806 that as a formula we get an+1 = an−1 + an ,
90 0.7855 which is the property defining the Fibonacci
100 0.7898 sequence.
200 0.8146
300 0.8269
500 0.8403 8.2 Infinite Series
1000 0.8589 ∞ k
ln (n!) 1
From the table, lim = 1. 1. 3 is a geometric series with
n→∞ ln (nn ) 5
k=0
When x → ∞, slope of the graph y = ln x ap- 1
proaches to 0. Due to this property, the given a = 3 and |r| = < 1, so it converges to
5
limit approaches to 1. 3 15
=
1 − 1/5 4
12 ∞
59. If side s = 12 and the diameter D = then 1 k
n 2. 5 is a geometric series with
3
the number of disks that fit along one side is k=0
12 |r| = 5 > 1, so it diverges.
12 = n. Thus, the total number of disks is
n ∞ k
12 12 1 1
· =n·n=n 2 3. − is a geometric series with
12 12 2 3
n n k=0
an = wasted area inbox with n2 disks 1 1
2 a = and |r| = < 1, so it converges to
6 2 3
= 12 · 12 − n2 π = 144 − 36π ≈ 30.9 1/2 3
n = .
1 − (−1/3) 8
∞ k
60. The answer is a5 = 30 or a5 = 31, depending 1
on the position of the points. 4. 4 is a geometric series with
2
k=0
1
a = 4 and |r| = < 1, so it converges to
2
4
= 8.
1 − 1/2
∞
1 k
5. (3) is a geometric series with
2
k=0
|r| = 3 > 1, so it diverges.
∞ k
1
6. 3 − is a geometric series with
2
k=0


1 13. Using partial fractions
a = 3 and |r| = < 1 so it converges to n
2 2k + 1
3 Sn =
= 2. k 2 (k + 1)2
1 − (−1/2) k=1
n
7. Using partial fractions, 1 1
= −
n
4
n
2 2 k2 (k + 1)2
k=1
Sn = = −
k(k + 2) k k+2 1 1 1 1 1
k=1 k=1 = 1− + − + −
2 2 2 2 4 4 9 9 16
= 2− + 1− + −
3 4 3 5 1 1
2 2 + ··· + 2
− 2
+ ··· + − (n − 1) n
n n+2 1 1
2 2 + −
=2+1− − n2 (n + 1)2
n+1 n+2
4n + 6 1
=3− 2 and =1−
n + 3n + 2 (n + 1)2
4n + 6 1
lim Sn = lim 3 − 2 = 3. and, lim Sn = lim 1− =1
n→∞ n→∞ n + 3n + 2 n→∞ n→∞(n + 1)2
Thus, the series converges to 3. Thus the series converges to 1.
4k ∞
8. lim = 4 = 0, so the series diverges by 4
k→∞ k + 2 14.
th
the k -Term Test for Divergence. k(k + 1)(k + 3)(k + 4)
k=1
∞
3k 4 1 1
9. lim = 3 = 0, so by the k th -Term Test = −
k→∞ k + 4
3 k(k + 4) (k + 1)(k + 3)
k=1
for Divergence, the series diverges. 4
∞
1 4
∞
1
= −
3 k(k + 4) 3 (k + 1)(k + 3)
10. This is a telescoping sum: k=1
∞
k=1
n n 41 1 1
9 3 3 = −
Sn = = − 34 k k+4
k(k + 3) k k+3 k=1
k=1 k=1
∞
3 3 3 3 3 3 41 1 1
= − + − + ··· + − − −
1 4 2 5 n n+3 32 k+1 k+3
3 3 3 3 3 k=1
=3+ + − − − . ∞
2 3 n+1 n+2 n+3 1 1 1
∞ = −
9 3 k k+4
k=1
Thus ∞
k(k + 3) 1 1 1
k=1 − −
3 3 3 3 6 k+1 k+3
= lim 3 + + − − k=1
n→∞ 2 3 n+1 n+2 1 1 1 1 1 1 1
3 = 1+ + + − +
− 3 2 3 4 6 2 3
n+3 5
3 3 11 = .
= 3+ + = . 9
2 3 2
∞ ∞ ∞ ∞
2 1 1 2
11. =2 and from Example 2.7, 15. is a geometric series with
k k k ek
k=1 k=1 k=1 k=2
∞ 2 1
1 a = 2 and |r| = < 1,so it converges to
diverges, so 2 diverges. e e
k 2 2
k=1 = 2 .
e 2 1− 1 e −e
e
∞ ∞
4 1
12. = 4 and the harmonic series
k+1 k 16. lim |ak | = lim 31/k = 30 = 1 = 0 , So by
k=0 k=1
∞ ∞ k→∞ k→∞
1 4 the k th Term Test for Divergence, the series
diverges, so diverges.
k k+1 diverges.
k=1 k=0


∞ ∞ ∞
1 1 1 1 gent if |r| = |2c + 1| < 1 .
17. − = − . The
2k k+1 2k k+1 That is, if −1 < 2c + 1 < 1 or −2 < 2c < 0
k=0 k=0 k=0
first series is a convergent geometric series but Therefore, the series converges for
the second series is the divergent harmonic se- −1 < c < 0.
ries, so the original series diverges.
∞ ∞ ∞
1 1 1 1 ∞
18. − k = − 2
2k 3 2k 3k 26. k
is a geometric series with a = 2
k=0 k=0 k=0 (c − 3)
1 1 1 k=0
= − = 1
1 − 1/2 1 − 1/3 2 and |r| = . The series converges, if
(c − 3)
∞ ∞ ∞ 1
2 1 2 1 |r| = < 1, that is |c − 3| > 1, there-
19. + k = + . c−3
3k 2 3k 2k
k=0 k=0 k=0 fore c > 4, c < 2. Thus the series converges for
The first series is a geometric series with
1 c < 2 and c > 4.
a = 2 and |r| = < 1 so it converges to
3
2
= 3. ∞
1 − 1/3 c
The second series is a geometric series with 27. is divergent for all values of c. Refer
1 k+1
k=0
a = 1 and |r| = < 1, so it converges to Excercies 12
2
1
= 2.
1 − 1/2
∞
2 1 28. Two cases possible.
Thus + k = 3 + 2 = 5.
3k 2 (i) c = 0: ⇒ given series is
k=0
∞ ∞
∞ ∞ ∞ 2
1 1 1 1 = 2 which is divergent
20. − k = − (ck + 1)
k 4 k 4k k=0 k=0
k=0 k=0 k=0
The second series is a convergent geometric se-
ries, but the first series is a divergent harmonic (ii) c = 0: ⇒ given series is
∞ ∞
series, so the series diverges. 2 2
=
(ck + 1) c k+ 1 c
3k k=0 k=0
21. lim |ak | = lim =3=0 ∞
k→∞ k→∞ k + 1 2 1
th
So by the k -Term Test for Divergence, the = 1
c k+ c
k=0
series diverges. ∞
1
k 3 Which is divergent as 1
22. The limit lim (−1)k 2 does not exists, so k=0
k+ c
k→∞ k +1 is divergent.
th
the series diverges by the k -Term Test for Di-
vergence.
23. Since k is positive integer, k is ration num-
5 29. The series appears to converge.
ber and can’t be multiple of π. Hence n 1
k n √
lim sin = 0. Therefore , the series di- k=1 k
k→∞ 5
verges. 2 1.250000000
4 1.423611111
π 8 1.527422052
24. The limit lim tan−1 (k) = = 0, so by the 16 1.584346533
k→∞ 2
k th Term Test for Divergence, the series di- 32 1.614167263
verges. 64 1.629430501
∞ 128 1.637152005
k 256 1.641035436
25. 3(2c + 1) is a geometric series with
k=0 512 1.642982848
a = 3 and |r| = |2c + 1| . The series is conver- 1024 1.643957981


5
1.6

1.5 4.5

1.4 4

1.3
3.5

1.2
3
0 20 40 60 80 100
1.1

1 32. The series appears to converge.
0 20 40 60 80 100

n
2k
n
k
k=1
30. The series appears to diverge.
n 1 1 2
n √ 2 4
k=1 k
3 5.333333333
2 1.707106781
4 6
4 2.784457050
5 6.266666667
8 4.371436800
6 6.355555556
16 6.663994609
7 6.380952381
32 9.941512173
8 6.387301587
64 14.60206410
9 6.388712522
128 21.21122790
10 6.388994709
256 30.57088534
512 43.81657304
1024 62.55526935 6

5
12

10
4

8
3

6

2
4
10 20 30 40 50

2
∞
10 20 30 40 50 33. (a) Assume ak converges to L.
k=1
Then for any m,
∞ m−1 ∞
31. The series appears to converge. L= ak = ak + ak
k=1 k=1 k=m
n ∞
1
n √ = Sm−1 + ak .
k=1 k k=m
2 4.500000000 ∞

4 5.125000000 So ak = L − Sm−1 ,
k=m
8 5.154836310
and thus converges.
16 5.154845485 ∞ ∞
32 5.154845485 (b) Since ak = (a1 + · · · + am−1 ) + ak
1024 5.154845485 k=1 k=m


∞1
and the finite sum a1 +· · ·+am−1 does not 37. Let Sn = . Then
affect convergence, both series converge or k=1 k
both diverge. 1 3 5
S1 = 1 and S2 = 1 + = . Since S8 > , we
2 2 2
34. By using partial fraction technique, we get: have
1 1 1
1 1 1 S16 = S8 + + + ··· +
1 =2 − 1 9 10 16
k k+ 2 k k+ 2 1 1 5 1
It may be observed that unlike example 2.3 > S8 + 8 = S8 + > + = 3.
16 2 2 2
1 1 So S16 > 3.
both and do not form the same se-
k k+ 1 1 1 1
2 S32 = S16 + + + ··· +
ries. Consequently, the terms in the nth par- 17 18 32
tial sum do not get canceled. Hence the said 1 1 1 7
> S16 + 16 = S16 + > 3 + =
method is not useful for this example. 32 2 2 2
7
∞ ∞ So S32 > .
2
35. Assume ak converges to A and bk con- If n = 64, then S64 > 4. If n = 256, then
k=1 k=1 S256 > 5. If n = 4m−1 , then Sn > m.
verges to B. Then the sequences of partial
sums converge, and letting 38. The nth partial sum of the series
n n
Sn = ak and Tn = bk , P = 1 − 1 + 1 − 1 + 1 · ·· is given by
n
k=1 k=1 k 1 n
we have lim Sn = A and lim Tn = B. Pn = (−1) = (1 + (−1) ).
n n
2
k=0
n
Here Sn = the nth partial sum of the series
Let Qn = (ak + bk ), the sequence of partial
1 − 1 + 1 − 1 + 1 · ··
k=1
∞
1 n
Which is =Sn = 1+ (1 + (−1) ).
sums for (ak + bk ). Since S, T , and Q are 2
n
k=1 k
As (−1) diverges, by theorem 2.3 the se-
all finite sums,
k=0
Qn = Sn + Tn . ries 1 + 1 − 1 + 1 − 1 + 1 · ·· diverges.
Then by Theorem 1.1(i),
A + B = lim Sn + lim Tn = lim(Sn + Tn ) For the series A = 1 + 1 − 1 + 1 − 1 + 1 · ··, nth
n n n n−1
∞ k
partial sum is given by, Sn = 1 + (−1) .
= lim Qn = (ak + bk ) k=0
n
∞
k=1
∞
Now, for the series
The proofs for (ak − bk ) and cak are P = 1 − 1 + 1 − 1 + 1 · ·· we can write:
P =1−1+1−1+1······
k=1 k=1
similar. 1 − P = 1 − (1 − 1 + 1 − 1 + 1 · · · · · ·) = P
∞ ∞ 1
36. Assume that ak converges and bk di- ⇒P =
2
k=1 k=1
∞
For the 1 + 1 − 1 + 1 − 1 + 1 · · · · · · we have:
verges. Now suppose that (ak + bk ) con- A = 1 + P = 1 − 1 + 1 − 1 + 1 · ··
k=1
verges. But, since 1 3
∞ ∞ ∞ =1+ =
2 2
bk = (ak + bk ) − ak The Cesaro sum
n
k=1 k=1 k=1 1
∞ lim Sk
n→∞ n
Theorem 2.3(i) implies convergence of bk . k=1
n
k=1 1 3
∞ = lim · lim Sk =0· =0
n→∞ n n→∞ 2
This contradiction shows that (ak +bk ) can- k=1
k=1
∞ ∞
not converge. The proof for (ak − bk ) is 39. (a) .9 + .09 + .009 + · · · = .9(.1)k
k=1 k=0
almost identical. which is a geometric series with a = .9


and |r| = .1 < 1 so it converges to This can’t be an integer since all the terms on
.9 the numerator are divisible by p except for the
= 1.
1 − .1 last term.
1 9 9
(b) 0.199999 = + + + ... 46. Step 0: [0, 1]
10 100 1000 Step 1: After removing middle third, set be-
∞ k
1 9 1 1 2
= + come 0, 3 ∪ 3 , 1
10 100 10 Step 2: After removing middle third from
k=0
9 1
each subinterval, set become 0, 9 ∪ 2 , 3 ∪
9 9
1 2 6 7 8
= + 100 = 9 , 9 ∪ 9, 1
10 1 − 1/10 10
18 18 All the end points of the subintervals belong
40. (a) 0.181818 = + + ... to Cantor Set. Therefore other four points are
100 10000
∞ 1 2 7 8 1
1 9 , 9 , 9 , 9 . Third term of the series is 4 27 .
= 18 . Sum of the series:
100k
k=1 1 1 1
This is a geometric series with = +2 +4 + ...
18 1 3 9 27
a = and |r| = , so the sum is 1 1 1
100 100 = +2 + 22 + ...
18/100 2 3 32 33
= .
1 − 1/100 11 1 1 1
134 134 = 1+2 + 22 + ...
(b) 2.134134 = 2 + + + ... 3 3 32
1000 1000000 1 1
∞
1 = 2
= 2 + 134 . 3 1− 3
1000k
k=1
=1
The second term is a geometric series with
134 1
a= and |r| = , so the sum is The length of the Cantor Set
1000 1000 = 1 − (Sum of the lengths of remove intervals)
134/1000 134
2+ =2 . = 1 − (Sum of the series)
1 − 1/1000 999
∞ ∞
=1−1
1 −1 =0
41. and
k k
k=1 k=1
47. We have from the proof of the Theorem 2.1,
42. Yes. 1 1 − xn+1
1+x+x2 +x3 +x4 +........+xn = .
1−x
43. The geometric series 1 + r + r2 + r3 + · · · con- Now consider
1 0 < x < 1 ⇒ 0 < xn < 1
verges to S = provided that
1−r − 1 < −xn < 0 or 0 < 1 − xn < 1
−1 < r < 1. But −1 < r implies that −r < 1,
1 1 1 − xn 1
and so 1 − r < 2, and therefore > , therefore 0 < <
1−r 2 1−x 1−x
1 1
which means that S > . ⇒ 0 < 1 + x + x2 + ... + xn <
2 1−x
44. Since ak converges, lim ak = 0, Now, for the interval −1 < x < 0
k→∞
1 0 < −x < 1
by Theorem 2.2. Therefore, lim does not 1<1−x<2
k→∞ ak
1 If n is even, −1 < xn+1 < 0. Then
exist. In particular, lim = 0. 0 < −xn+1 < 1
k→∞ ak
1 ⇒ 1 < 1 − xn+1 < 2
Therefore, diverges, by the converse of 1 1 − xn+1 2
ak ⇒ < <
Theorem 2.2. 1−x 1−x 1−x
1 − xn+1
45. Suppose p is a prime, Therefore, 1 + x + x2 + ... + xn = >
1−x
1 1 1 p! + p!/2 + · + p!/p 1
1+ + +·+ =
2 3 p p! 1−x


Thus, the inequality does not hold if n is even. uct on the left side of the inequality. This
product will yield the sum of the terms of the
1
∞ form ,where k is either a prime or a product
1 k
48. The series is a geometric series with of the primes or the product of a prime and a
2k composite or it is the product of two compos-
k=0
1 ites, so that we can write the left side of the
a = 1 and r = , its sum is 2.
2 1 1 1 1
∞ inequality as 1 + + + + · · · · + ,where
1 2 3 4 n
Therefore, = 2. n is a positive integer greater than the largest
2k
k=0 prime p.
Now, the nth partial sum of this series is given Thus, we have
by 1 1 1 1 3 5 p
n
1 1
n 1+ + + +····+ < 2· · ··········
=2 1− < 2 2 3 4 n 2 4 p−1
2k 2 It may be observed that, if there would have
k=0
1 1 been finite number of primes then there would
Therefore 1 + + ........ + n < 2 ...(i) have been finite number of terms on either side
2 2
∞ of the inequality, which is not the case hence
1
The series is a geometric series with we conclude that there are infinite number of
3k
k=0 primes.
1 3
a = 1 and r = , its sum is . n−1
3 2 L
∞ 49. The amount of overhang is .
1 3 2(n − k)
Therefore, = k=0
3k 2 7
k=0 L
th So if n = 8, then = 1.3589L.
Now, the n partial sum of this series is given
2(8 − k)
by k=0
n
1 3 1
n
3 When n = 4,
= 1− < n−1 3
3k 2 3 2 1 1
k=0 L =L
1 1 3 2(n − k) 2(4 − k)
k=0 k=0
Therefore, 1 + + ........ + n < ...(ii)
3 3 2 = 1.0417L > L
∞
1 n−1 n
The series is a geometric series with L L
5k lim = lim
k=0 n→∞ 2(n − k) n→∞ 2k
1 5 k=0 k=1
a = 1 and r = , its sum is . n
5 4 L 1
∞ = lim =∞
1 5 2 n→∞ k
Therefore, = k=1
5k 4
k=0 n n
th
Now the n partial sum of this series is given 1 1 1
50. = −
by k(k − 1) k−1 k
k=2 k=2
n n
1 4 1 5 1 1 1 1 1 1
= 1− < = − + − + ··· + −
5k 5 5 4 1 2 2 3 n−1 n
k=0 1
1 1 5 = 1 − , we have
Therefore, 1 + + ........ + n < ...(iii) n
5 5 4 ∞
In general we can write for a largest prime p, 1 1
= lim 1 − = 1.
1 1 p k(k − 1) n→∞ n
1 + + ........ + n < k=2
p p p−1 On the other hand,
Now, multiplying the inequalities ∞ ∞
1 1
(i),(ii),(iii).etc. We get, k =
k(k − 1) (k − 1)
1 1 1 1 k=2 k=2
1+ +····+ n 1+ +····+ n ∞
2 2 3 3 1
= = ∞.
1 1 1 1 k
1+ +····+ n ··· 1+ +····+ n k=1
5 5 p p 2
3 5 p 51. p2 + 2p(1 − p)p2 + [2p(1 − p)] p2 + · · ·
∞
< 2· · ·········· k
2 4 p−1 = p2 [2p(1 − p)] is a geometric series with
for any positive integer n. Consider the prod- k=0


a = p2 and |r| = 2p(1−p) < 1 because 2p(1−p) The sum represents the probability that you
is a probability and therefore must be between eventually win a game.
0 and 1. So the series converges to
p2 p2
= .
1 − [2p(1 − p)| 1 − 2p + 2p2
If, p = .6, 2v
∞
.62 2v g 2v
1 − 2(.6) + 2(.6)2
= .692 > .6. 58. rk = =
g 1−r g(1 − r)
k=0
1 p2 2v
If p > , > p. ∞
2 1 − 2p(1 − p) 2v g 2v
r2k = =
1 1 1
∞ k g 1 − r2 g(1 − r2 )
k=0
52. Since 1 + + 2 + ··· =
12 12 12
k=0
1 12
= = , the minute hand and the hour
1 11
1− 59. (a) Take a number less than 1000 and try.
12
hand are in the same location approximately 5
minutes and 27 seconds after 1.
(b) The series
∞ c c c
53. d + de−r + de−2r + · · · = d(e−r )k c + 6 + 12 + 18 + .......... is an in-
10 10 10
k=0 finite geometric series with a = c and
which is a geometric series with 1
a = d and e−r = e−r < 1 if r > 0. r= 6 <1,
10
∞ c
k d it converges to
So d(e−r) = . 1 − 10−6
1 − e−r
k=0 where
If r = .1, x 999−x
c 103 + 106
−6 =
∞
d 1 − 10 1 − 10−6
d(e−.1 )k = =2
1 − e−.1 1000x + 999 − x 999x + 999
k=0
so d = 2(1 − .905) ≈ .19 = =
999999 999999
∞ k x+1
15 1 =
54. 30 + 15 + + ··· = 30 1001
2 2
k=0
30
= = 60 miles. (c) Let x + 1 be a positive inte-
1 − 1/2 ger(assumed) less than 1000 such that
The bikes meet after 1 hour. In that time, a
x+1
fly flying 60 mph will have traveled 60 miles. /11 /13 =
7
∞ k
3 100,000 x+1
55. 100,000 = = $400,000. =
4 1 − 3/4 1001
k=0 c c c
= c + 6 + 12 + 18 + .....
56. P = ce−r + ce−2r + ce−3r + . . . 10 10 10
∞ ∞
k x 999 − x x 999 − x
= ce−kr = c e−r = 3+ 6 + 9+
10 10 10 1012
k=1 k=1
x 999 − x
ce−r c + 15 + +···
= = r 10 1018
1 − e−r e −1 First three digits are given by the term
x
57. Since 0 < p < 1, therefore −1 < −p < 0, and only. Therefore the first three dig-
0 < 1 − p < 1. Thus, the given series is geo- 103
x+1
metric with common ratio r = 1 − p, and so its of are the digits of x. Thus
103
converges: here the digits of the number on the left
∞ ∞
side repeat after every 6 digits and first
p(1 − p)n−1 = p rn−1
three digits ( i.e x) is less by one than the
n=1 n=1
1 p p assumed number (x + 1) and last three
=p = = = 1. digits are 9’s complements of x.
1−r 1 − (1 − p) p

Ism et chapter_8

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