Thông tin tài liệu
Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ CHAPTER 10 10.1 INFINITE SEQUENCES AND SERIES SEQUENCES a1 121 0, a2 122 14 , a3 123 92 , a4 124 16 1, a a1 1! 2! 12 , a3 ( 1)2 3! 16 , a4 ( 1)3 4! 24 ( 1) ( 1)5 a1 21 1, a2 41 13 , a3 61 15 , a4 81 71 a1 (1)1 1, a2 (1)2 3, a3 (1)3 1, a4 (1)4 a1 22 12 , a2 a1 21 22 23 12 , a2 22 1 22 34 , a3 a1 1, a2 12 23 , a3 32 a9 511 , 256 , 362,880 a10 12 , 23 1 23 24 25 24 1 24 78 , a4 74 , a4 74 23 15 16 15 , a5 15 8 31 , a 16 24 63 , 32 255 , a7 127 , a8 128 64 ( 1)2 (2) 1, a9 128 , a10 1( 2) a9 92 , a4 16 , 24 a5 241 , 120 a6 , 720 a7 , 5040 a8 , 40,320 3,628,800 a1 2, a2 , a8 64 22 12 , a4 a10 1023 512 a1 1, a2 12 , a3 a9 23 24 12 , a3 a3 ( 1)3 (1) 12 , a4 1, ( 1)4 12 a5 1, ( 1)5 14 , a , a6 16 32 256 10 a1 2, a2 1, a3 a8 14 , a10 15 2( 1) 32 , a4 1, 3 23 a5 2, 4 12 5 a6 13 , a7 72 , 11 a1 1, a2 1, a3 2, a4 3, a5 5, a6 8, a7 13, a8 21, a9 34, a10 55 12 a1 2, a2 1, a3 12 , a4 12 , 1 a5 12 1, 12 a6 2, a7 2, a8 1, a9 12 , a10 13 an (1)n 1 , n 1, 2, 14 an (1)n , n 1, 2, 15 an (1)n 1 n , n 1, 2, 16 an ( 1)n 1 n2 , n 1, 2, Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 701 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 702 Chapter 10 Infinite Sequences and Series 17 an 2n 1 , 3( n 2) 18 an n2(nn51) , n 1, 2, n 1, 2, 19 an n 1, n 1, 2, 20 an n 4, n 1, 2, 21 an 4n 3, n 1, 2, 22 an 4n 2, n 1, 2, 23 an 3n , n! 25 an 1 ( 1)n 1 , n 1, 2, 27 n 1, 2, 29 lim 1 n lim n 1 2n n 30 lim 2n 1 lim lim 31 15n 4 n n 8n lim 32 lim n 3 n n 5n lim 33 n n 1 n 1 n lim 34 lim 1n n 70 n 38 lim n 2 n 1, 2, n n 5 4 n n 1 n 5 converges n3 n ( n 3)( n 2) ( n 1)( n 1) n 1 n n 2 lim n n 70 n2 n n lim converges lim (n 1) diverges n diverges 36 lim (1) n 1n does not exist diverges n n2n1 1 1n nlim 1 1n 12 converges 2 n lim n 21 n , converges lim (1)n does not exist diverges n n 12 ( 1)n diverges 3 n n lim n 1, 2, (Theorem 5, #4) 1n 2 lim 2 1 1n n n 13 n lim n3 , 5n 1 ( 1)n lim 1 n converges n n ( 1)n n n 37 26 an lim (0.1)n converges n 28 35 24 an 2n 3 converges 2n 39 ( 1) n 1 n n 1 lim converges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.1 Sequences 40 n 41 lim 42 43 44 45 46 lim 12 n 2n n n 1 lim 10 n n n (0.9) lim lim converges 2n n 2n n 1 n ( 1)n lim n lim 21 converges n n diverges lim sin 2 1n sin lim 2 1n sin 2 converges n n lim n cos (n ) lim (n )(1)n does not exist diverges n n lim sin n n lim sin n 2n n n because 47 lim nn n lim 48 3n n n lim 49 ln ( n 1) n n lim 50 ln n n ln 2n 1n n 22n lim n n ln 3n ln n 3n lim lim sin n n n sin n 2n 2n because 1n n converges by the Sandwich Theorem for sequences converges by the Sandwich Theorem for sequences ˆ converges (using l'Hopital's rule) 3n (ln 3)2 6n n lim n11 n lim n n n 1 lim 3n (ln 3)3 n lim lim n ˆ diverges (using l'Hopital's rule) converges n 1 1n converges lim 81 n converges (Theorem 5, #3) lim (0.03)1 n converges (Theorem 5, #3) 53 lim 7n n n e7 converges (Theorem 5, #5) 54 n n ( 1) lim 1 n e1 converges n 51 52 55 n n lim 1n lim n n n 10n lim 101 n n1 n 1 converges n (Theorem 5, #5) (Theorem 5, #3 and #2) Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 703 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 704 56 57 Chapter 10 Infinite Sequences and Series lim n lim n n 12 converges n n n lim 31 n 1n lim n3 n n 11 converges lim n n (Theorem 5, #2) (Theorem 5, #3 and #2) n 58 59 60 61 62 lim (n 4)1 ( n 4) lim x1 x converges; (let x n 4, then use Theorem 5, #2) n lim x ln n 1n n n lim n1 n 1 diverges (Theorem 5, #2) lim ln n ln (n 1) lim ln nn1 ln lim nn1 ln converges n n n lim n 4n n lim n n 1 converges n lim n n n 1 n 64 ( 4) n n n ! 65 lim n6!n n 10 lim n 66 lim nn !n n 67 1 n n n lim exp n n n! n (Theorem 5, #6) diverges lim exp ln n ln ln lim 1n n 6n2 (3n 1)(3n 1) (Theorem 5, #6) n ln e converges 3n 1 3n 1 exp (Theorem 5, #6) ln 1 ln n exp ln n e1 converges n1 nlim lim exp n ln n (Theorem 5, #3) lim 1n and nn! lim nn! converges n n n n diverges n n n n! 3n 1 n n 3n 1 lim 106 lim lim ln 1n n 123( n 1)( n ) nnnnn lim (ln n ) lim n converges lim (Theorem 5, #2) lim 32 (1 n ) lim 32 31 n 1 converges n lim nn! n n 69 n n 63 68 lim ln n 3 ln (3n 1) ln (3n 1) 3n 1 3n 1 lim exp lim exp n n n n2 e2 converges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ (Theorem 5, #5) Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.1 Sequences 70 n lim nn1 n n n 1 lim exp n ln n 705 1 ln n ln ( n 1) lim exp lim exp n n 1 n n n 12 n lim exp n (nn 1) e1 converges n 71 1/ n xn n n 1 lim lim x n 2n11 1/ n x lim exp n 2 ln(2n 1) exp exp 1n ln 2n11 x nlim x nlim n 2n xe0 x, x converges 72 lim n n2 n lim exp n ln n n2 ln 1 1 lim exp n lim exp n n lim exp n n n n n2 e0 converges 73 n n lim 3 n6 n n ! n 74 1011 n n n 11 10 12 lim 36n ! converges n 1211 1011 n n n n n 12 12 11 11 10 11 12 n 75 lim (Theorem 5, #6) n 120 121 0 n n 108 1 110 n n lim lim n e e n n e 1 n 2e n n ln n ln n lim sinh (ln n) lim e 2e lim n diverges n n n 77 lim 79 80 (Theorem 5, #4) n n 2n 2n lim n lim en e n lim e2 n 1 lim 2e2 n lim converges 76 78 converges n sin 1n 2n 1 n lim sin n1 n n n2 lim cos n n n2 lim 3 n n n n1 2 n2 cos converges 1cos 1n lim sin n n2 lim sin converges lim n cos 1n lim n n n n n 1 lim n n sin lim n n 1n lim n n n sin 2 n n lim n n cos n n n3 2 n3 lim cos n cos converges n 3n ln 3 5n ln ln 3n 5n n n n n n lim exp ln exp lim exp 1 nlim n n n 3n ln 3 ln 3n n lim exp ln 3 ln exp(ln 5) lim exp n n 1 n n 35 1 5n Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 2 n n 1 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 706 81 Chapter 10 Infinite Sequences and Series lim tan 1 n 2 converges 82 n lim tan 1 n 2 converges n n 83 n n n lim 13 lim 13 converges n 2n n 84 ln n n n n lim exp n lim exp 22n 1 e0 converges n n n n n 85 (ln n)200 n n lim n lim ln n 86 87 88 lim n n 200 (ln n )199 n n lim 200199 (ln n )198 n n 5 ln n 4 n 10 ln n lim lim lim n n n n n lim n 1 n n lim n lim 200! converges n lim lim n n2 n lim n n2 n n n n n n (Theorem 5, #4) n2 1 n n n2 n n2 n n 80 ln n n lim n n n n n2 n lim n n n n 1 n2 n n 1 lim 3840 n converges n 1 1 1n lim n 1 n n 1 n lim converges 1 n 1 n1 n2 1n 1 2 converges 89 n1 n n x 90 n ln n dx lim n lim 1n converges n n n xp dx lim 11 p p11 lim 11p p11 p11 if p converges n x n n lim lim n (Theorem 5, #1) 72 n 1 an 91 Since an converges lim an L lim an 1 lim n n L 172 L(1 L) 72 L L2 L 72 L 9 or L 8; since an for n L 92 Since an converges lim an L lim an 1 lim n n an n an L L6 L2 L( L 2) L L2 L L 3 or L 2; since an for n L 93 Since an converges lim an L lim an 1 lim n n n 2an L L L2 L L 2 or L 4; since an for n L 94 Since an converges lim an L lim an 1 lim n n n 2an L L L2 L L 2 or L 4; since an for n L Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.1 Sequences 5an L 5L L2 5L L or L 5; 95 Since an converges lim an L lim an 1 lim n n 707 n since an for n L 96 Since an converges lim an L lim an 1 lim 12 an L 12 L L2 25L 144 n n n L or L 16; since 12 an 12 for n L 97 an 1 a1 , n 1, a1 Since an converges lim an L lim an 1 lim a1 L L1 n n n n n L2 L L 2; since an for n L 98 an 1 an , n 1, a1 Since an converges lim an L lim an 1 lim an L L n n n L2 L L 12 ; since an for n L 12 99 1, 1, 2, 4, 8, 16, 32, … 1, 20 , 21 , 22 , 23 , 24 , 25 , x1 and xn 2n 2 for n 2 100 (a) 12 2(1)2 1, 32 2(2)2 1; let f (a, b) a 2b a b a 4ab 4b 2a 4ab 2b 2b a ; a 2b 1 f (a, b) 2b2 a 1; a 2b f (a, b) 2b a 1 (b) rn2 2 a 2b a b 2 a ab 4b2 a ab 2b2 a b 2 In the first and second fractions, yn n Let a b a 2b a b 1 r yn2 n 2 a 2b a b n f ( x) x 2; the sequence converges to 1.414213562 (b) f ( x) tan ( x) 1; the sequence converges to 0.7853981635 4 (c) f ( x) e x ; the sequence 1, 0, 1, 2, 3, 4, 5, … diverges (b) (c) (d) lim f x lim n x n x x 1 1 lim n tan n f (0) 1, 1 n lim n f f 0x f (0) x f (0), where x f ( x) tan 1 x lim n e1/ n f (0) e0 1, f ( x) e x n lim n ln 2n f (0) 1 2(0) 2, f ( x) ln (1 x) n Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ a b and b n and a b 2b 2n n yn n Thus, lim rn 102 (a) yn represent the (n 1) th fraction where for n a positive integer Now the nth fraction is 101 (a) n Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 708 Chapter 10 Infinite Sequences and Series 2 103 (a) If a 2n 1, then b a2 n 24n 1 2n 2n 12 2n2 2n, c a2 2n 2n 12 4n 8n3 8n 4n 2n 2n 1 c 2n 2n and a b 2n 1 2n 2n (b) 104 (a) a2 2 lim a a lim 2n2 2n n n 1 a (2 n) lim 2n or lim exp n n a2 2 lim a a ln 2n 2n Stirling’s approximation n n ! n n (b) 40 50 60 105 (a) lim ln n c n n 1n c 1 n cn nc a 2 21n e0 1; n! ne n 2n , for large values of n n e c n cn lim lim sin lim sin ne 2n 1 (2n) ne 14.71517765 18.39397206 22.07276647 0 (b) For all 0, there exists an N e(ln ) nc 1 4n2 4n 4n 8n3 4n 2 lim exp 22n lim exp n n n! 15.76852702 19.48325423 23.19189561 lim nc c such that n e(ln ) c n n lim c ln n ln c ln nc ln 1 0 106 Let {an } and {bn } be sequences both converging to L Define {cn } by c2 n bn and c2 n 1 an , where n 1, 2, 3, For all there exists N1 such that when n N1 then an L and there exists N such that when n N then bn L If n max{N1 , N }, then cn L , so {cn } converges to L 107 lim n1 n lim exp 1n ln n lim exp 1n e0 n n n 108 lim x1 n lim exp 1n ln x e0 1, because x remains fixed while n gets large n n 109 Assume the hypotheses of the theorem and let be a positive number For all there exists an N1 such that when n N1 then an L an L L an , and there exists an N such that when n N then cn L cn L cn L If n max{N1 , N }, then L an bn cn L bn L lim bn L n 110 Let We have f continuous at L there exists so that x L f ( x) f ( L) Also, an L there exists N so that for n N , an L Thus for n N , f (an ) f ( L) f (an ) f ( L ) Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.1 Sequences 709 3( n 1) 1 111 an 1 an ( n 1) 1 3nn11 3nn24 3nn11 3n 3n 4n 3n2 6n n 2; the steps are reversible so the sequence is nondecreasing; 3nn11 3n 3n 3; the steps are reversible so the sequence is bounded above by 2( n1) 3! (2n 3)! (2n5)! (2n 3)! (2n5)! ( n 2)! 112 an 1 an ( n 1)! ( n 2)! ( n 1)! (2n 3)! ( n 1)! (2n 5)(2n 4) n 2; ( n1) 1! the steps are reversible so the sequence is nondecreasing; the sequence is not bounded since (2 n 3)! ( n 1)! (2n 3)(2n 2) (n 2) can become as large as we please 2n 13n 1 ( n 1)! n 1 n 1 n n ( n 1)! n3! n3n n ! n which is true for n 5; the steps are reversible so the sequence is decreasing after a5 , but it is not nondecreasing for all its terms; a1 6, a2 18, a3 36, 113 an 1 an a4 54, a5 324 64.8 the sequence is bounded from above by 64.8 114 an 1 an n21 n11 n2 1n n2 n21 2 2n 1 2n n ( n21) 2n11 ; the steps are reversible so the sequence is nondecreasing; 2n 1n the sequence is bounded from above 115 an 1n converges because 1n by Example 1; also it is a nondecreasing sequence bounded above by 116 an n 1n diverges because n and 1n by Example 1, so the sequence is unbounded 117 an 2n 1 2n 1 118 an 2n 1 3n 1n ; since 1n (by Example 1) a nondecreasing sequence bounded above by 2n 23 n and 3n 2n 2n 0, the sequence converges; also it is ; the sequence converges to by Theorem 5, #4 119 an (1)n nn1 diverges because an for n odd, while for n even an 1n converges to 2; it diverges by definition of divergence 120 xn max {cos 1, cos 2, cos 3, , cos n} and xn 1 max {cos 1, cos 2, cos 3, , cos (n 1)} xn with xn so the sequence is nondecreasing and bounded above by the sequence converges 121 an an 1 1 2n n 1 2( n 1) n 1 n 2n 2n n 2n 2n n n and thus the sequence is nonincreasing and bounded below by 1 n n 2; it converges ( n 1) 1 122 an an 1 nn1 n 1 n 2n n 2n and nn1 1; thus the sequence is nonincreasing and bounded below by it converges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 710 123 Chapter 10 Infinite Sequences and Series 4n 1 3n 4n 4 34 n so an an 1 34 n 4 34 n 1 34 34 n n 1 1 and 34 n 4; thus the sequence is nonincreasing and bounded below by it converges 124 a1 1, a2 3, a3 2(2 3) 22 22 3, a 22 22 23 23 3, a5 23 23 3 24 24 3, , an 2n 1 2n 1 2n 1 2n 1 2n 1 (1 3) 2n 3; an an 1 2n 2n 1 2n 2n 1 so the sequence is nonincreasing but not bounded below and therefore diverges 125 For a given , choose N to be any integer greater than / Then for n N , sin n 1 sin n 0 n n n N 126 For a given , choose N to be any integer greater than 1/ e Then for 1 n N , n n N 127 Let M and let N be an integer greater than 1MM Then n N n 1MM n nM M n M nM n M (n 1) nn1 M 128 Since M1 is a least upper bound and M is an upper bound, M1 M Since M is a least upper bound and M1 is an upper bound, M M1 We conclude that M1 M so the least upper bound is unique ( 1)n 129 The sequence an is the sequence 12 , 32 , 12 , 32 , This sequence is bounded above by clearly does not converge, by definition of convergence 130 Let L be the limit of the convergent sequence {an } Then by definition of convergence, for corresponds an N such that for all m and n, m N am L 3, but it there and n N an L 2 Now am an am L L an am L L an 2 2 whenever m N and n N 131 Given an 0, by definition of convergence there corresponds an N such that for all n N , L1 an and L2 an Now L2 L1 L2 an an L1 L2 an an L1 2 L2 L1 2 says that the difference between two fixed values is smaller than any positive number 2 The only nonnegative number smaller than every positive number is 0, so L1 L2 or L1 L2 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 ak ( n) and ( n) , where ak ( n ) L1 , ( n) L2 and L1 L2 Thus ak ( n ) ( n ) L1 L2 So there does not exist N such that for all m, n N am an So by Exercise 128, the sequence {an } is not convergent and hence diverges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.1 Sequences 711 133 a2 k L given an there corresponds an N1 such that 2k N1 a2 k L Similarly, a2k 1 L 2k N a2 k 1 L Let N max{N1 , N } Then n N an L whether n is even or odd, and hence an L 134 Assume an This implies that given an there corresponds an N such that n N an an an an an On the other hand, assume an This implies that given an there corresponds an N such that for n N , an an an an an xn x xn xn a x2 a xn2 a n f ( x) x a f ( x) x xn 1 xn 2n x xn 1 xn xn n (b) x1 2, x2 1.75, x3 1.732142857, x4 1.73205081, x5 1.732050808; we are finding the positive 2 135 (a) a number where x 0; that is, where x 3, x 0, or where x 136 x1 1, x2 cos (1) 1.540302306, x3 1.540302306 cos (1 cos (1)) 1.570791601, x4 1.570791601 cos (1.570791601) 1.570796327 2 to decimal places After a few steps, the arc xn 1 and line segment cos xn 1 are nearly the same as the quarter circle 137-148 Example CAS Commands: Mathematica: (sequence functions may vary): Clear[a, n] a[n_ ] : n1/ n first25 Table[N[a[n]],{n, 1, 25}] Limit[a[n], n 8] The last command (Limit) will not always work in Mathematica You could also explore the limit by enlarging your table to more than the first 25 values 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, the following Clear[minN, lim] lim Do[{diff Abs[a[n] lim], If[diff 01, {minN n, Abort[]}]}, {n, 2, 1000}] minN For sequences that are given recursively, the following code is suggested The portion of the command a[n_ ]: a[n] stores the elements of the sequence and helps to streamline computation 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}] Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 712 Chapter 10 Infinite Sequences and Series The limit command does not work in this case, but the limit can be observed as 1.25 Clear[minN, lim] lim 1.25 Do[{diff Abs[a[n] lim], If[diff 01, {minN n, Abort[]}]}, {n, 2, 1000}] minN 10.2 INFINITE SERIES sn sn sn a 1 r n (1 r ) a 1 r n (1 r ) 21 1 lim sn 21 1 n 1 a 1 r n (1 r ) n 100 100 1 1 1 100 n 12 n 9 lim sn 1001 11 1 100 n lim sn n 32 1 ( 2) n sn 1( 2) , a geometric series where r divergence ( n 1)( n 2) n ( n 1) n11 n 1 sn sn 12 12 13 13 14 n11 n1 12 n1 nlim sn 25 35 35 54 n51 5n 5n n51 n51 nlim n5 n51 sn 25 , the sum of this geometric series is 14 16 64 16 , the sum of this geometric series is 64 256 , the sum of this geometric series is 16 64 1 14 1 12 1 14 161 1 14 12 74 1 14 , the sum of this geometric series is 10 54 16 64 11 (5 1) 1 14 4 52 13 45 19 85 271 , is the sum of two geometric series; the sum is 1 13 10 32 23 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.2 Infinite Series 52 13 45 91 85 271 , is the difference of two geometric series; the sum is 12 (5 1) 1 12 13 (1 1) 1 12 10 23 17 1 13 12 15 14 251 18 1251 , is the sum of two geometric series; the sum is 1 15 56 17 10 16 ; the sum of this geometric series is 14 45 25 125 25 125 1 15 Series is geometric with r Converges to 1 25 16 Series is geometric with r 3 3 Diverges 17 Series is geometric with r 18 Converges to 1 18 18 Series is geometric with r 23 23 Converges to 19 0.23 23 100 102 n n 0 21 0.7 n 0 23 0.06 n 10 10 23 100 23 1 100 99 101 106 101 n n0 24 1.414 414 1000 103 n 25 1.24123 26 3.142857 n0 n 0 n d n 10 10 n 234 1000 234 1 1000 999 10d d 1 10 1006 1 1 10 90 15 414 1000 414 1413 999 999 1 1000 142,857 106 106 n0 1 123 105 103 234 1000 103 n0 22 0.d n 0 124 100 25 1 23 20 0.234 107 1 10 23 n 124 100 123 5 10 1 13 10 3 124 100 142,857 106 1 16 10 3 123 105 102 142,857 106 1 124 123 100 99,900 3,142,854 999,999 116,402 37,037 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 123,999 99,900 41,333 33,300 713 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 714 Chapter 10 Infinite Sequences and Series 27 lim n n n 10 28 n ( n 1) lim ( n 2)( n 3) lim 2n n lim 22nn15 lim n n n 5n n n 29 lim n n 30 lim 2n n n 3 31 32 33 34 n lim diverges 2 diverges test inconclusive n n lim test inconclusive lim cos 1n cos diverges n en n e n n lim n n lim ne lim en lim n e 1 n e n diverges lim ln 1n diverges n lim cos n does not exist diverges n sk lim 1 k11 1, 12 13 13 14 k11 1k 1k k11 k11 klim k 35 sk 12 series converges to 36 sk 13 34 34 93 93 163 (k 31) k2 3 k ( k 1) ( k 1)2 lim sk lim 3, series converges to ( k 1) k k 37 sk ln ln ln ln ln ln ln k ln k ln k ln k ln k ln ln k lim sk lim ln k ; series diverges k k 38 sk tan1 tan tan tan1 tan tan tan k tan k 1 tan k 1 tan k tan k 1 tan tan k 1 lim sk lim tan k 1 does not exist; series diverges k k 12 cos1 13 cos1 13 cos1 14 cos1 14 cos1 15 cos 1 1k cos 1 k11 cos 1 k11 cos 1 k 1 3 cos 1 k 1 lim sk lim 3 cos 1 k 1 3 2 6 , series converges to 6 k k 39 sk cos 1 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.2 Infinite Series 40 sk 5 6 715 k 3 k 2 k 4 k 3 k 4 2 lim sk lim k ; series diverges k k 41 (4 n 3)(4n 1) 15 91 91 131 4k17 4k13 4k13 4k11 4k11 n13 n11 sk 15 lim sk lim k11 k 42 k A(2 n 1) B (2n 1) 2nA1 nB1 (2 n 1)(2 n 1) A(2n 1) B(2n 1) (2 A B )n ( A B) 6 (2 n 1)(2 n 1) k 2 A B A B A A and B 3 Hence, A B A B n 1 3 43 1111 1 1 3 5 40 n (2 n 1)2 (2 n 1)2 A (2 n 1) 2( k 1) 1 k 1 k 1 B (2 n 1)2 (2 nC1) D (2 n 1) (2n 1)(2n 1) 31 the sum is k 1 k 3 n 1 2n11 2n11 lim 2k11 k A(2 n 1)(2 n 1) B (2n 1)2 C (2 n 1)(2 n 1)2 D (2n 1) (2n 1)2 (2 n 1) A(2n 1)(2n 1)2 B(2n 1) C (2n 1)(2n 1)2 D(2n 1)2 40n A(8n3 4n2 2n 1) B (4n 4n 1) C (8n3 4n 2n 1) D(4n 4n 1) 40n (8 A 8C )n3 (4 A B 4C D )n2 (2 A B 2C D )n ( A B C D) 40n A 8C AC A B 4C D A B C D BD B 20 B 2 A B 2C D 40 A B C D 20 B D 20 A B C D A B C D AC and D 5 C and A Hence, A C k k 1 1 1 40 n 1 1 (2n 1)2 (2 n 1)2 (2 n 1)2 (2 n 1)2 25 25 2( k 1) 12 (2k 1)2 (2 k 1)2 n 1 n 1 1 the sum is lim (2 k 1) (2 k 1) n 44 n 1 n ( n 1)2 12 n ( n 1)2 sk 14 14 19 19 16 ( k 1)2 k k ( k 1)2 lim sk lim 1 ( k 1) k k lim 1 1 45 sk lim sk k 2 k k 1 k k k 1 1 k 1 k 1 46 sk 12 11 11 11 11 11 (1k 1) 11 k 11k (1k 1) 12 (1k 1) 2 2 2 2 2 lim sk 12 11 12 k Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 716 Chapter 10 Infinite Sequences and Series 47 sk ln13 ln12 ln14 ln13 ln15 ln14 ln (1k 1) ln1k ln ( k1 2) ln (1k 1) ln12 ln ( k1 2) lim sk ln12 k 48 sk tan 1 (1) tan 1 (2) tan 1 (2) tan 1 (3) tan 1 (k 1) tan 1 (k ) tan 1 (k ) tan 1 (k 1) tan 1 (1) tan 1 (k 1) lim sk tan 1 (1) 2 4 2 4 k 49 convergent geometric series with sum 1 2 1 2 50 divergent geometric series with r 51 convergent geometric series with sum 52 32 1 12 lim (1)n 1 n diverges n n 53 The sequence an cos starting with n is 1, 0, 1, 0,1, 0, 1, 0, , so the sequence of partial 2 sums for the given series is 1, 1, 0, 0, 1, 1, 0, 0, and thus the series diverges 54 cos (n ) (1) n convergent geometric series with sum 55 convergent geometric series with sum 56 lim ln n 3n 1 12 e 58 convergent geometric series with sum 1 101 1 1x 59 difference of two geometric series with sum lim 1n n e2 e2 1 diverges 57 convergent geometric series with sum 60 1 15 n lim n1 n n 2 20 18 9 x x1 1 23 1 13 32 e1 diverges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.2 Infinite Series 61 63 n! n n 1000 diverges lim n 1 2n 3n 4n 2n 4n n 1 n 1 converge since r n 1 64 2n 3n 4n n n lim 2n 4n n 12 34 ; both 12 n n 1 lim n n 1 n 1 and r n nnn n 12n lim and 34 n 1 1, respectively n 1 12 n n lim n diverges n are geometric series, and both 1 12 and 34 n n 1 1 34 3 by Theorem 8, part (1) lim n 65 3n 4n nn n n ! 62 717 2n 4n 1 3n 4n 1 12 1 n n 1 n lim diverges by n th term test for divergence nn1 ln(n) ln(n 1) ln n 1 n 1 sk ln(1) ln(2) ln(2) ln(3) ln(3) ln(4) ln(k 1) ln(k ) ln(k ) ln(k 1) ln(k 1) lim sk , diverges k 66 lim an lim ln n n 2nn1 ln 12 diverges 67 convergent geometric series with sum 68 divergent geometric series with r 69 n0 70 n0 n 0 n0 23.141 22.459 1 n 0 e (1)n x 2n x e e e (1)n x n ( x) n ; a 1, r x; converges to 71 a 3, r 72 1 x 1 ; ; a 1, r x ; converges to converges to 1 x21 12 3sin1 x n ( 1)n 3 sin x for all x (since n n 0 3sin x 73 a 1, r x; converges to 1 ( x ) n 3 x for 1 ; a 12 , r x1 11x for x 1 1 x for x or 1 x 1 ; 3sin x converges to 12 1 3sin1 x for all x) 1 2x for x or x Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 3 sin x 2 sin x sin x 832sin x Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 718 Chapter 10 Infinite Sequences and Series 74 a 1, r x2 1 21 x ; converges to 75 a 1, r ( x 1); converges to 3 x ; 76 a 1, r converges to 1ln x 78 a 1, r ln x; converges to 79 (a) n 2 80 (a) n 1 (c) x2 for or x 2 x for x or 2 x x 1 for 3 x or x for x (2k 1) 2 , k an integer for ln x or e1 x e ( n 4)( n 5) (b) ( n 2)( n 3) (b) (c) ( n 2)( n 1) (c) n0 n 3 ( n 2)( n 3) n 5 ( n 3)( n 2) n 20 ( n 19)( n 18) 12 1 12 3 3 one example is 32 34 83 16 1 12 1 one example is 12 14 81 16 1 12 81 (a) one example is (b) 32 x 1sin x 77 a 1, r sin x; converges to x2 x 1 1 ( x 1) 1 82 The series 14 81 16 n 1 k 12 is a geometric series whose sum is n0 k2 k 1 12 where k can be any positive or negative number 83 Let an bn 12 n 84 Let an bn 12 n 85 Let an 86 Yes: 14 n a1n Then Then and bn n 1 n 1 n 1 n 1 n 1 n 1 an bn 12 n an bn 12 12 n Then A diverges The reasoning: 1, while bnn (1) diverges n 1 n 1, while a n 1 n 1 n 1 (anbn ) 14 13 AB an 13 , B bn and bnn 12 n 1 n 1 n 1 an converges an an nth-Term Test Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ n a n n 1 A B diverges by the an Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.2 Infinite Series 719 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 An a1 a2 an and lim An A Assume n an bn converges to S Let Sn a1 b1 a2 b2 an bn Sn a1 a2 an b1 b2 bn b1 b2 bn Sn An lim b1 b2 bn S A bn converges This contradicts the bn assumption that 1 r 89 (a) 5 n diverges; therefore, r r 35 ; 53 53 132 13 r r ; 13 13 (b) 1 r 10 90 eb e2b 10 1 eb 9 an bn diverges 2 103 132 103 eb eb 89 b ln 13 103 89 91 sn 2r r 2r r 2r r n 2r n 1 , n 0, 1, sn r r r n 2r 2r 2r 2r n 1 lim sn if r n 1 r 2r 1 r 1 22r , 1 r or r 92 area 22 2 (1)2 2 12 m2 1 12 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 300 e( 0.12)(24) 94 L sn 1ar a 1 r n 1 r ar n 300 e( 0.12)(24) e( 0.12)(24) 17.84 mg n 1 r 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 27 27 3 9 (b) The lengths of the intervals removed are: Stage 1: 1 Stage 2: 1 3 Stage 3: 2 and so on 1 27 Thus the sum of the lengths of the intervals removed is 2 n 1 n 1 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 1 (2 / 3) Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 720 Chapter 10 Infinite Sequences and Series 96 (a) n 1 n 1 L1 3, L2 43 , L3 43 , , Ln 43 lim Ln lim 43 n n s , (b) Using the fact that the area of an equilateral triangle of side length s is , A A 3(4) A A 3(4) , A A 3(4) , , A 3(4) 3(4) A2 A1 43 n 2 4 n 33 k 2 4 32 3 3 k 2 k 1 4k 9k 1 34 n k 2 12 k 3 k 1 3 364 1 x 1 x2 3 , 27 n k 3 3 k 1 k 2 201 43 1 35 43 85 58 A1 n 1 n 1 n2 4 b x 0.2 b dx lim b dx lim 54 x0.8 1 b dx diverges 10.2 diverges n 1 n is positive, continuous, and decreasing for x 1; b x 0.2 1 0.2 lim 12 tan 1 2x lim 1 b b b dx lim 1x b n x x 4 b x2 b dx lim f ( x) 10.2 is positive, continuous, and decreasing for x 1; f ( x) 21 x2 1 dx converges 12 converges b0.8 54 1 x1 b 1 dx x2 4 b dx b x lim 12 tan 1 b2 12 tan 1 12 4 12 tan 1 12 1 x 14 dx converges converges f ( x) x 1 is positive, continuous, and decreasing for x 1; dx x4 lim ln | b 4| ln 5 b 3 32 f ( x) 12 is positive, continuous, and decreasing for x 1; lim , THE INTEGRAL TEST b 3 n lim 1b 12 k 2 lim An lim n n 10.3 we see that A1 diverges n 1 n4 f ( x) e2 x is positive, continuous, and decreasing for x 1; b lim 12 e2 x lim 12b 12 1 b 2e 2e b 2e2 dx x4 1 b lim ln x b diverges 2 x 1 e dx lim b 2 x b 2 n e2 x dx converges e Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ b dx x 4 b lim n 1 e dx converges Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.3 The Integral Test x (ln x )2 f ( x) b lim ln1x lim ln1b ln12 b b f ( x) x x2 x 3; 3 ln x x n 3 f ( x) x2 ex 7 x (ln x )2 b lim 2 diverges n n2 4 n 1 f ( x) dx lim n 1 b ln x x b dx converges n n2 4 x2 x 4 lim b 15 82 n 3 n n2 n (ln n )2 dx converges for x 2, thus f is decreasing for b b b x (ln x ) dx lim n 2 ln b 4 2 ln(13) x dx x2 diverges is positive and continuous for x 1, f ( x) n 2 b x b e dx lim 327 e7 13 e x4 x x 1 x3 lim ln x b n 2 x2 ex n2 en ln n n diverges e 327 e7 23 x 4 ( x 1)2 n4 n n 1 b x 1 8 ln ln n n n 3 x ( x 6) 3e x b x2 ex e1 16 e4 dx converges n 7 25 e5 362 e n7 n2 en n2 en for x 6, thus f is decreasing for x 7; 3b 18b 54 eb 327 lim 73 e b x4 ( x 1)2 2 14 16 25 36 n 8 1 11 converges; a geometric series with r 10 n 4 n2 n 1 x4 ( x 1) 7 x ( x 1)3 327 e7 for x 7, dx diverges n 8 n4 n n 1 diverges n n n 1 dx x 1 5ln(n 1) 5ln Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 12 converges; a geometric series with r 1e 13 diverges; by the nth-Term Test for Divergence, lim n dx x 1 eb b b b b dx lim x 1 dx dx lim x11 dx dx ( x 1) ( x 1)2 b ( x 1) b b 1 3( 6b 18) converges lim ln b b31 ln 73 14 diverges by the Integral Test; converges is continuous for x 2, f is positive for x 4, and f ( x) 8 dx diverges diverges lim dx lim 3xx 18x x3 54 e e x b b e thus f is decreasing for x 8; ln x x b dx lim ln x lim 2(ln b) 2(ln 3) b b ln n n 54 b3 b e lim 10 x (ln x )2 2 f ( x) lnxx is positive and continuous for x 2, f ( x) 2ln2 x for x e, thus f is decreasing for x 3; x 3 ln x x dx blim b x x dx x 4 n 3 ln is positive and continuous for x 1, f ( x) diverges 2 is positive, continuous, and decreasing for x 2; 721 diverges Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 722 Chapter 10 Infinite Sequences and Series 15 diverges; n n 1 16 converges; n 1 3 n n 1 2 n n , which is a divergent p-series with p 2 n 1 n3 2 , which is a convergent p-series with p 17 converges; a geometric series with r 18 18 diverges; n 1 8 n 8 n 1 n and since n n 1 19 diverges by the Integral Test: 2 lnxx dx 12 ln 20 diverges by the Integral Test: 2 tet dt lim 2tet ln b n ln x x 5n n n 3 23 diverges; n0 2 n 1 n0 , n 1 2n n n 1 25 diverges; lim an lim n n ln n lim n n dx x , dx e dt t n which diverges by the Integral Test 12 ln(2n 1) as n 2n ln n lim n dx x x 1 lim n n n n n u x 1 n 1 du ; ln dx u du x 1 28 diverges; lim an lim 1n n ln x dx x 1 ln n ln 4 n dx x 1 26 diverges by the Integral Test: 27 diverges; lim lim 1 24 diverges by the Integral Test: n n ln b 5n ln n n ln diverges 4et lim 2eb (b 2) 2e(ln 2) (ln 2) ln b lim 2 n n 1 dx; t ln x, dt 21 converges; a geometric series with r 22 diverges; lim diverges, 8 e 29 diverges; a geometric series with r ln 1.44 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ n ln as n Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.3 The Integral Test 30 converges; a geometric series with r du lim ln u u 1 b 1 1 0.91 1x 3 31 converges by the Integral Test: ln 723 (ln x ) (ln x ) 1 u ln x, du x dx dx; b sec1 u lim sec1 b sec1 (ln 3) lim cos1 ln b b b1 sec1 (ln 3) cos (0) sec (ln 3) 2 sec 1 (ln 3) 1.1439 1u 1x 1 x11ln x dx 1 32 converges by the Integral Test: b dx; u ln x, du 1x dx 1 (ln x )2 du lim tan 1 u lim tan 1 b tan 1 2 2 b b 1n lim sin x x 0 x 33 diverges by the nth-Term Test for divergence; lim n sin 1n nlim sin 34 diverges by the nth-Term Test for divergence; lim n tan tan n n sec lim sec 1n n e 1u u e ex 1 e x n dx; 1 e x b lim ln uu1 lim e b b 37 converges by the Integral Test: 38 diverges by the Integral Test: 1 tan 1 x 1 1 39 converges by the Integral Test: 1 x x dx; x 1 1 u e x , du e x dx, dx du du u e u (1u ) ln bb1 2ln ee1 ln1 ln ee1 2ln ee1 0.63 u tan 1 x dx; 8u du 4u dx du 1 x u x2 du x dx x x Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ b 2 4 16 34 lim 12 ln u lim 12 (ln b ln 2) b b tan e dx blim b 1 e sech x dx lim b du 2 u lim tan 1 eb tan 1 e tan 1 e 0.71 b dx; u e x , du e x dx 1 b u21 du n 1n lim n2 sec n 1n n 12 du lim tan 1 u lim tan 1 b tan 1 e 2 tan 1 e 0.35 e b n 36 converges by the Integral Test: lim 35 converges by the Integral Test: n n 1 x b e 1 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ 724 Chapter 10 Infinite Sequences and Series 40 converges by the Integral Test: 1 sech x dx lim b b sech x dx lim x 1 lim (tanh b 1) b b b 0.76 41 b (b 2) a ln x ln x lim ln b 1 xa x 1 dx blim b a ln lim ln 3a b (b 2) a b4 ln ; 3a , a (b 2) lim b a lim (b 2)a 1 the series converges to ln 53 if a and diverges to if 1, a b b a If a 1, the terms of the series eventually become negative and the Integral Test does not apply From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges a 42 3 2a x 1 x 1 lim b 12 a b (b 1) dx lim ln b x 1 ( x 1)2 a lim a 1 b a (b 1) b b 1 blim ln (b 1)2 a ln 3 1, a , a 2 lim ln 42 a b the series converges to ln b 1 (b 1)2 a ln ; 42 a 42 ln if a 12 and diverges to if a 12 If a 12 , the terms of the series eventually become negative and the Integral Test does not apply From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges 43 (a) (b) There are (13)(365)(24)(60)(60) 109 seconds in 13 billion years; by part (a) sn ln n where n (13)(365)(24)(60)(60) 109 sn ln (13)(365)(24)(60)(60) 109 ln(13) ln(365) ln(24) ln(60) ln(10) 41.55 44 No, because n 1 45 Yes If an nx x n 1 n and 1n diverges n 1 is a divergent series of positive numbers, then n 1 12 an a2 also diverges and a2 an n 1 n 1 n There is no “smallest” divergent series of positive number: for any divergent series numbers 2n a n an n 1 has smaller terms and still diverges n 1 Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ of positive ... n 36 converges by the Integral Test: lim 35 converges by the Integral Test: n n 1 x b e 1 Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file... bounded below by it converges Copyright 2014 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file... https://TestbankDirect.eu/ n a n n 1 A B diverges by the an Solution Manual for Thomas Calculus Multivariable 13th Edition by Thomas Full file at https://TestbankDirect.eu/ Section 10.2
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