Solution manual for thomas calculus multivariable 13th edition by thomas

CHAPTER 10 INFINITE SEQUENCES AND SERIES... Assume the hypotheses of the theorem and let  be a positive number... Let k n and i n be two order-preserving functions whose domains are t

CHAPTER 10 INFINITE SEQUENCES AND SERIES

n   does not exist  diverges 36 lim ( 1) 1n 1n

n   does not exist  diverges

1 1

n n

1 lim

n n n n

lim exp n n n exp



3 2 2

cos sin

3 3

ln 3 ln 5 ln 3 ln 5

1 1

n

n n

n n

95 Since a n converges lim n lim n 1 lim 5 n 5 2 5 0 0

2

n n

In the first and second fractions, y n  Let n a

b represent the (n th fraction where 1) a b and 1 b n  1for n a positive integer 3. Now the nth fraction is a a b2b and a b 2b2n  2 n y n n

Thus, lim n 2

n r

 

101 (a) f x( )x2 the sequence converges to 1.4142135622;  2

(b) ( ) tan ( ) 1;f x  x  the sequence converges to 0.7853981635 4

(c) ( )f x e x; the sequence 1, 0, 1, 2, 3, 4, 5, … diverges

      ( )f x e x 1

1 2(0)lim ln 1 n (0) 2,

      ( ) ln (1 2 )f x   x

2 2

2 2

lim lim sin lim sin 1

a a

     because x remains fixed while n gets large

109 Assume the hypotheses of the theorem and let  be a positive number For all  there exists an N1 such that when n N 1 then a n    L   a n    L  L  a n, and there exists an N2 such that when n N 2then c n    L   c n  L  c n L  If nmax{ ,N N1 2}, then L  a n b nc n  L 

111 3( 1) 1 3 1 3 4 3 1 2 2

a  a            n  n n  n  n n    the steps are reversible so the sequence is nondecreasing; 3 1n n1  3 3n 1 3n   3 1 3;

the steps are reversible so the sequence is bounded above by 3

2 n n   the sequence is bounded from above 2

115 a n  converges because 1 1n 1n by Example 1; also it is a nondecreasing sequence bounded above by 1 0

116 a n   diverges because n   and n 1n 1n by Example 1, so the sequence is unbounded 0

  the sequence converges; also it is

a nondecreasing sequence bounded above by 1

118 2 1  2 1

3

3n n n 3n;

n

a     the sequence converges to 0 by Theorem 5, #4

119 a n ( 1)n1  n n1 diverges because a n  for n odd, while for n even 0 a n 2 1 1n converges to 2; it diverges by definition of divergence

120 max {cos 1, cos 2, cos 3,x n  , cos }n and x n1max {cos 1, cos 2, cos 3,, cos (n1)}x n with x n 1

so the sequence is nondecreasing and bounded above by 1 the sequence converges

n n

thus the sequence is nonincreasing and bounded below by 2 it converges

122 a na n1n n1 ( 1) 1n n 1 n22n 1 n22n  and 1 0 n n1 thus the sequence is nonincreasing 1;and bounded below by 1 it converges

123 4 1 3  3

4

n n n

125 For a given , choose N to be any integer greater than 1 /  Then for nN,

128 Since M1 is a least upper bound and M2 is an upper bound, M1M2 Since M2 is a least upper bound and 1

M is an upper bound, M2M1 We conclude that M1M2 so the least upper bound is unique

129 The sequence a n  1 ( 1)2n is the sequence 12 2 2 2, , , ,3 1 3  This sequence is bounded above by 3

2, but it clearly does not converge, by definition of convergence

130 Let L be the limit of the convergent sequence { } a n Then by definition of convergence, for 2 there

corresponds an N such that for all m and n, m N  a m L 2 and n N  a n L 2 Now

132 Let ( )k n and ( )i n be two order-preserving functions whose domains are the set of positive integers and whose ranges are a subset of the positive integers Consider the two subsequences a k n( ) and a i n( ), where

( ) 1,

k n

a L a i n( )L2 and L1L2 Thus a k n( )a i n( )  L1L2  So there does not exist N such that 0

for all m, n N  a ma n  So by Exercise 128, the sequence { }a n is not convergent and hence diverges

133 a2k   given an L 0 there corresponds an N1 such that 2kN1 a2k  L  Similarly,

a   L  k N  a   L  Let Nmax{ ,N N1 2} Then n N  a n L  whether n

is even or odd, and hence a n L

134 Assume a n This implies that given an 0 0 there corresponds an N such that n N  a n 0 

         On the other hand, assume a n  This implies that given 0

an 0 there corresponds an N such that for n N , a n   0  a n   a n   a n  0 0

number where x2  that is, where 3 0; x23,x or where 0, x 3

136 x1 1, x2 1 cos (1) 1.540302306, x31.540302306 cos (1 cos (1)) 1.570791601,  

4 1.570791601 cos (1.570791601) 1.570796327 2

x     to 9 decimal places After a few steps, the arc 

x n1 and line segment cosx n1 are nearly the same as the quarter circle

137-148 Example CAS Commands:

Mathematica: (sequence functions may vary):

If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01

of the limit, do the following

Clear[a, n]

a[1] 1;

a[n ] : a[n] a[n 1] (1/5)_     n 1

first25 Table[N[a[n]],{n, 1, 25}]

The limit command does not work in this case, but the limit can be observed as 1.25

1 12 1 

9 74167 647 , the sum of this geometric series is  

 

7 1 4

7 3 1 

10 5 54 165 645 , the sum of this geometric series is  1

12     5 1 5 1 5 1 

2 3 4 9 8 27(5 1)       , is the difference of two geometric series; the sum is

16 Series is geometric with r      Diverges 3 3 1

17 Series is geometric with r 18 18   Converges to 1 11

8

1 7 1 

18 Series is geometric with r      Converges to 23 23 1

 

2 3 2

2 5 1

1000

1000 10 1 999 0

1

00

3 10

123,999 41,333

100 10 10 100 1 100 10 10 100 99,900 99,900 33,300 0

n    test inconclusive

3lim n lim n 0

2 1

2 1



50 divergent geometric series with r  2 1

51 convergent geometric series with sum  

 

3 2 1 2

55 convergent geometric series with sum 22

1

1

1 1

e

e e

61 !

1000lim n n 0

n     diverges 62 lim n n n! lim 1 2n n n n lim

n n





 and  3

4 1

n n

1 3 4

n n n

3 sin

3 sin 3 sin

8 2sin

2 4 sin 1

x

x x

74 1, 12;

x

a r  converges to 22

1 2

1

1 1

x

x x

n n

a b

87 Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series

88 Let A na1a2  a n and lim n

93 (a) After 24 hours, before the second pill:300e( 0.12)(24) 16.840 mg;after 48 hours, the amount present

after 24 hours continues to decay and the dose taken at 24 hours has 24 hours to decay, so the amount present is 300e( 0.12)(48) 300e( 0.12)(24) 0.945 16.840 17.785 mg. 

(b) The long-run quantity of the drug is   ( 0.12)(24)

( 0.12)(24)

( 0.12)(24) 1

a r

L s      

95 (a) The endpoint of any closed interval remaining at any stage of the construction will remain in the Cantor

set, so some points in the set include 0,27 27 9 9 27 27 3 3 9 91 , 2, , ,1 2 7 , 8, , , , , 1.1 2 7 8 (b) The lengths of the intervals removed are:

Stage 1: 1

3 Stage 2: 1 1 1 2

4 4

n n n

n n n

n

n n n

11 converges; a geometric series with r101  1 12 converges; a geometric series with r  1e 1

13 diverges; by the nth-Term Test for Divergence, lim n n1 1 0

n   

14 diverges by the Integral Test; 51 51

1n x dx5ln(n 1) 5ln 2 1x dx 

  which diverges by the Integral Test

24 diverges by the Integral Test: 1

30 converges; a geometric series with rln 31 0.91 1

31 converges by the Integral Test:  1

32 converges by the Integral Test:

   

1 2 2

40 converges by the Integral Test: 2 2  

a If a the terms of the series eventually become negative and the Integral Test does not apply From 1,that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges

1 ( 1) 2 ( 1)

43 (a)

(b) There are (13)(365)(24)(60)(60)  109 seconds in 13 billion years; by part (a) s n  1 lnn where





 has smaller terms and still diverges