Unformatted text preview: 366 Section 9.1 8. (a) We graph the points n, 1
1 2n
for n
2n 1, 2, 3, … . 2. (a) Note that a0 1
,a
32 1, a1 1n
.
3 an 1
, and so on. Thus
9 1, a2 1
,a
23 (c) Note that a0 5, a1 0.5, a2 5(0.1)n 5
.
10n (b) Note that a1
( 1)n
n an 1 1
, and so on. Thus
3 . [0, 23.5] by [ 2, 2] 1
n→ 1 (b) lim an 2n
2n lim n→ 2
2 1 an
1
for n
n 9. (a) We graph the points n, 2 1, 2, 3, … . 0.05, and so on. Thus 1n1
alternate
2 3. Different, since the terms of ∑ n1 between positive and negative, while the terms of
1n1
are all negative.
2 ∑ n1 4. The same, since both series can be represented as [0, 23.5] by [ 1, 3] (b) lim an 1
n lim 2 n→ n→ 10. (a) We graph the points n, 1
2 1 2
ln (n 1)
for n
n 1, 2, 3, … . 1
4 …. 1
8 5. The same, since both series can be represented as
1
2 1 1
4 …. 1
8 6. Different, since ∑ ∑
n [0, 23.5] by [ 1, 1]
1 ln (n 1)
lim
n
n→ (b) lim an
n→ lim n→ (n 1) 1
u4 and 1
u1 1, 1
, so
16 u1 1, u2 un n2, or * ( 1)
2n 1 1 2n
3 1 1
8 4, u3 11
,
4 u3 1
u2 9, and u4 1
,
9 9. Converges; ∑ n0 5
4 2n
3 5
4 1 5
4 3 2
3 15
4 10. Diverges, because the common ratio is 1 and the terms do not approach zero.
11. Diverges, because the terms alternate between 1 and
and do not approach zero. 16. We may write n2. 12. Converges; ∑ 3( 0.1)n 1 n0 3
( 0.1) 30
11 3 … (b) Let un represent the value of * in the nth term, starting
with n 0. Then 1
u3 u3 1
u0 1, 1
, so u0
16 1, u1 16. We may write un (c) If *
( 1)3
( 1)6 11
,
4 u2 1
u1 4, u2
(n 1
1 1
4 2 ( 1)4 n0 1)2, or * (n 1)2. 1
2 ( 1)5 1
3 3. 1 1 1 2 2(
(2 2 1 2 1 1 2 n0 1 2
n 1 2 1 2 ∑ 2 n 4 9, and … , which is the same as the desired series. Thus let * 13. Converges; ∑ sin n 1
, and
9 3, the series is
2 … but 3 2
3 1 1
4 8. Diverges, because the terms do not approach zero. 1. (a) Let un represent the value of * in the n th term, starting
1. Then 1 n0 Section 9.1 Exercises with n n 7. Converges; ∑ 0 1 n1 1n1
1
1
2
2
1
1
1
….
2
4
8 2 2 2
1)( 2 1) 2
1) 2
2 1 2 2 1 14. Diverges, because the terms do not approach zero. 1 Section 9.1 e 15. Converges; since 1, ∑ 0.865 16. Converges; ∑ n0 6n 17. Since ∑ 2 x n0 n0 1 1 function f (x) 1 2x 18. Since ∑ ( 1)n(x 1 2x ∑[ 1)n n0 (c) The partial sums alternate between positive and
negative while their magnitude increases toward
infinity. 1
.
2 x 1)]n, the series (x 24. Since ∑ n0 converges when (x 1) 1 1
, the series represents the
1 tan x
1
,
k
x
k.
4
4
tan x (b) The partial sums are alternately 1 and 0. , the series represents the 1
2 , k , where k is any integer. Since the 4 23. (a) Since the terms are all positive and do not approach
zero, the partial sums tend toward infinity. 11
, . Since
22 1 and the interval of convergence is the sum of the series is x function f (x) n n0 k sum of the series is 1 5
6 1 n0 1. Thus, the series converges for 4 ∑ (2x) , the series converges when nn 2x 1 tan x 1
6 15n
66 ∑ 5n ∑ (tan x)n, the series converges when n0 e 1 n1 e 22. Since ∑ tan n x 1 en ∑ en
ne n0 1 and the interval of n e
e , this is a geometric series with n0 e common ratio r 1.03, which is greater than one. e convergence is ( 2, 0). Since the sum of the series is
1
[ (x 1 1
1)]
1 f (x) x 2 x , 2 2 25. , the series represents the function x ∑ xn 20 n0 1 0.
1 20, x x 1
1n
(x
2 19. Since ∑ n0 converges when ∑ 3)n 2 n0 3 x
2 xn 3 20 20x 19
20 1 and the interval of convergence 1 x 1 . 2 2 the series represents the function f (x) x) x 1 ,1 x 5. For any real number a
use a
2 a
4
1
2 use 1
20. For ∑ 3
n0 x 1n x
2 a
16
1
8 ∑ 2r n 1…
.
32 1 1 3 x 5r , the series 3 x , 1 x Series: ∑ 2 3. n1 21. Since ∑ sin x
n n0 sin x ∑ (sin x) , the series converges when
n n0 1. Thus, the series converges for all values of x except odd integer multiples of 2 , that is, x for integers k. Since the sum of the series is
series represents the function f (x)
(2k 1) .
2 3
5 1 1
1
,
sin x (2k 1) (b) ∑ n1 3n1
5
13
2 13 n 1
r
2 1 r
13
2 1 5 5r 5r
r
Series:
n1 13
2 5r 5, r 2 1
, the
sin x 1 3 2 6 represents the function f (x) 1) 5 r 6 3
(x 1: 5, r r 2 Since the sum of the series is x … . To get 0, a
32
1
16 2 1 n1 1 and the interval of convergence is ( 1, 3). 2 a
8
1
4 0, 27. Assuming the series begins at n , the series converges when (a) 1 20x 26. One possible answer: 2 1
(3 1 19 x , the series is (1, 5). Since the sum of the series is 367 3n1
10 3
2
3
10 368 Section 9.1 28. Let a
0.21 21
and r
100 0.21 1
, giving
100 33. 1.414 0.21(0.01)2 0.21(0.01) 1 0.414 0.414(0.001)2 0.414(0.001) ∑ 0.414(0.001)n 1 n0 … 0.21(0.01)3 ∑ 0.21(0.01)n 1 n0 157
111 0.21
1 0.01
0.21
0.99
7
33 29. Let a
0.234 34. 1.24123 1
, giving
1000 0.234(0.001) 0.234(0.001)3 0.234(0.001)2 … 124
100
41,333
33,300 0.234
1 0.001
0.234
0.999
26
111 0.7(0.1) 35. 3.142857 0.7(0.1)2 0.7(0.1)3 3
3
3 36. Total distance …) 0.142857 ∑ 0.000001n
(0.142857) 1 1
0.000001 0.142857
0.999999
1
7 0.06(0.1) ∑ 0.06(.1)n 4(0.6)2 2[4(0.6) 4 2 4(0.6)3 …] 2 ∑ 2.4(0.6)n 4 …] 4 4 0.13 26 n0 ∑ 0.06
1 0.1
0.06
0.9
1
15 0.000001 22
7 d
[1 0.1 0.12
10
d
(0.1)n
10 n 0
d
1
10 1 0.1
d1
10 0.9
d
9 n0 0.142857(1 n0 … 0.7
1 0.1
0.7
0.9
7
9 0.06 3 3 ∑ 0.7(0.1)n 32. 0.06 … 0.00123
1 0.001
n1
0.00123
0.999
41
33,300 0.0000012 n0 31. 0.d 0.00123(0.001) ∑ 1.24 ∑ 0.234(0.001)n 0.7 0.00123 1.24 n0 30. 0.7 1.24 0.00123(0.001)2 234
and r
1000 0.234 0.414
1 0.001
46
111 1 1 2.4
0.6 16m 37. Total time
0.06(0.1)2 0.06(0.1)3 4
4.9 … 4(0.6)2
4.9 4(0.6)
4. 9 2
4(0.6)3
4.9 … 4
4.9 2 4(0.6)
[1
4. 9 4
4.9 2 4(0.6)
4.9 7.113 sec 0.6
1
1 0.6 ( 0.6)2 …] 369 Section 9.1
38. The area of each square is half of the area of the preceding
square, so the total of all the areas is ∑ 4
1 8 m 2. 1
2 1n
2 n1 n1 ∑ 2 ∑ 45. Comparing a
a S a ar n
1r 41. Using the notation Sn ar a 1
4 1, then lim r n
n→ ar 2 ar ar 3 1 r 4) , the first term is a (x 4).
… 4)2 (x a ar n
1r
n→ a
1 r ∑ ar n 4 1 3x a
1 r the common ratio is r
3x 9x 1 … 4)n 1, so the 1) with a
1 r , the leading term is a ( 3x) .
1 and (x 1
(x
4 1) … 1)2 x
2x with the common ratio is r
2x 2 4x 3 3x 1
2 x as (x 1
(x 1 1) a
r , the first term is a 11
,.
22 1, so the 2n 1 x n a
1 1 and the common ratio is r
… r x , … 1) (x 1) Rewriting 1
2 2 (x 1) 1 n 1. 1, so the 1
1
x
21 and comparing with x
2
1
and the common ratio is r
2 the first is a
1
2 as 1
x
4 12
x
8 … a
1 r , x
.
2 … 1n
x
2n 1 x and 48. 1 eb e 2b e 3b … x
2 1, so the interval ∑ (e b)n n0 1 Interval: The series converges when 2x
of convergence is 1 … of convergence is ( 2, 2). 2x.
… x. 1)n and comparing with 1, so the 11
,.
33 1 1 1
( 1)n(x
4 Interval: The series converges when 1 1) interval of convergence is (0, 2). Series: … n Interval: The series converges when
interval of convergence is , the first term is Alternate solution: 3x.
… 2 1
(x Interval: The series converges when x 1 1, then the nth partial sum is na, which goes to
with 1
(x
4 Series: 1 . a ar n
has no finite limit and
1r
n 1 1
41 The first term is a 1 n1 diverges. Series: x ( 1)n(x 1
and the common ratio is r
4 47. Rewriting 1, then r n has no finite limit as n → , so the expression 43. Comparing 1 ar n 1, ... a ar
.
1r 0 and so ∑ ar n lim 1 or r Series: 1 a with interval of convergence is (0, 2). the formula from Exercise 40 is Sn 42. Comparing 4) Interval: The series converges when x n If r 1, so the interval Series: n rS S (1 n→ (x 46. Comparing r: ar n r) S If r … 3x 3n interval of convergence is (3, 5). (a ar ar 2 ar 3 ... ar n 2 ar n 1)
(ar ar 2 ar 3 ar 4 ... ar n 1 ar n)
a ar n a lim Sn … Interval: The series converges when x (b) Just factor and divide by 1 If r 3 and x 3. 3x 6 1
(x 1 Series: 1
1
2 2 rS , the first term is a and the common ratio is r 4 40. (a) S r of convergence is ( 1, 1). 1n
42 1 a
1 Interval: The series converges when x 3 12
2n 1
2 n0 3x 3 Series: 3 ∑ 2n 39. Total area with x3 1 the common ratio is r 2 n0 4 3 44. Comparing 1n …
1, so the interval 1
eb 9 1 9 9e b 8 eb 8
9 b ln 9e b 8
9 ln 8 ln 9 370 Section 9.1 49. (a) When t n0 1 1 t. x (1
1 t t 1 t, which is true when t For t 1
.
2 (1 ∑ n0 2(1 n t
1 10 when t 1 1 t t 1 1 (1 t
t) 1 t t, so 4
1 t2 with a
1 r , the first term First four terms: 4 4t
n 4t (1 4t 2n 0, so the constant term of the power series for G (x) will be 0. Integrate the terms for f (x) to
obtain the terms for G (x).
43
45
47
x
x
x
3
5
7
4
General term: ( 1)n
x 2n 1
2n 1 First four terms: 4x (c) The series in part (a) converges when 2 x) C. 4x 3 … nx n x) d
(1
dx 2 2x 3x 2 4x 3 … 3 x) 0
2 2 2 …) 12x 2 6x n (n 1)x n ... … 12x 2 6x 1)x n n (n (n …. 1)x n 2)(n 1, so the interval (b) No, because if you integrate it again, you would have
the original series for f, but by Theorem 2, that would
have to converge for 2 x 2, which contradicts
the assumption that the original series converges only
for 1 x 1. t2 1, so the lim an. Then by definition of convergence, for n→ there corresponds an N such that for all m and n,
n, m N ⇒ am L 2 and an L 2 Now,
1, which result in the am an 4 4
3 4
5 4
7 … ( 1)n 4
2n 1 … and
G ( 1)
4 4
3 respectively. 4
5 4
7 … ( 1)n 4 1 2n 1 …, whenever m am L am convergent series
G (1) …. 53. (a) No, because if you differentiate it again, you would
have the original series for f, but by Theorem 1, that
would have to converge for 2 x 2, which
contradicts the assumption that the original series
converges only for 1 x 1. 54. Let L interval of convergence is ( 1, 1).
(d) The two numbers are x 2 … 2 2, this may be written as
12x 2 6x 1 … 1 of convergence is ( 1, 1). 6 General term: ( 1) (4t )
(b) Note that G (0) dx Interval: The series converges when x t 2. 4 and the common ratio is r
4 3 3x 2 Replacing n by n t 2 x) 2x 1 Thus, f (x) f (x) is a 2 x) 9. 50. (a) Comparing f (t) … 4 nx n 1
, we have
2 (c) For t 2(1 d
(1
dx 1
.
2 Thus, S converges for all t 1)4 (x 1
dt
t 1 Using the result from Example 4, we have: t, which is always true. S 1)2
(x 1)3
2
3
( 1)n(x 1)n
n
(x 1 f (x) dx
1 (x 52. To determine our starting point, we note that 0, the inequality is equivalent to 0, the inequality is equivalent to t x 1) 1)2 … 1)3 (x 1)n t), which is always false. For (x … , we may write ln x 1 1, this inequality is equivalent to t S 1
x ( 1)n(x 1, or t t 51. Since 2. 1
2 1 t (b) S converges when
For t 1 1n
2 ∑ 1, S L L
N and n an an L
N. 2 2 = . 2 ...
View
Full
Document
This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.
 Spring '08
 ALL

Click to edit the document details