Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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72 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 11.7 Strategy for Testing Series 1. -- 1 < - 1 = (1)n - for all n > 1. n+3n f (~)n is a convergent geometric series [lrl = t < 1), so f ~ 3n 3 - n = l . n=l n + 3 converges by the Comparison Test. 3. lim la..l = lim _2:_ 2 = 1, so lim an= lim (-1}"_2:_ does not exist. Thus, the series f (- 1)n __.!:_diverges by n - oo n-oo n + n--
17 lim I an+l I= SECTION 11.7 STRATEGY FOR TESTING SERIES 0 73 lim 11 · 3 · 5 .. · · · (2n- 1)(2n 1) . 2 · 5 · 8 · .... (3n - 1) I = lim 2n 1 · n-oo a n n--+oo 2 · 5 · 8 · · · · · (3n- 1)(3n + 2) 1 · 3 · 5 · · · · · (2n- 1) n--+oo 3n + 2 = lim 2 + 1/ n = ~ < 1 , n--+oo 3 + 2/ n 3 oo 1 . 3. 5 .. · .. (2n - 1) so the series 2::: ( 3 ) converges by the Ratio Test. n=l 2 · 5 · 8 · · · · · n - 1 . ln x 1 2 - ln x ? lrin. . 2 19. Let f(x) = r . Then f (x) = 2 x 312 < 0 when lnx > 2 or x > e-, so r ts decreasmg for n > e . vx . . vn . I R I lim In n I' 1/n 1' 2 0 h . ~ ( )"Inn b By I'Hosptta 's u e, r = tm ( ) = tm r = , sot e senes n6- 1 - 1 r;;, converges y the n--+oo vn n--+oo 1/ 2 Vn n - oo vn - vn Alternating Series Test. 21 . lim lan. l = lim IC- l )ncos(l/n 2 )1 = lim icos(1/n 2 ).1 =cos O= 1, so the series E (-l)'' cos(l/ n 2 ) diverges by the n-oo n-oo n -'oo . n =l Test for Divergence. 23. Using the Limit Comparison Test with a n = tan(.!.) and b,. = .!., we have n . n lim an = lim tan(1/ n) = lim tan(l/ x) M: lim sec2(1/x). (- 1/x2) = lim sec2(1/ x) = 12 = 1 > 0. Since n-oo bn n --+oo 1/ n :z:--+oo 1/x :r-oo -1/x 2 :z:--+oo 00 00 2::: bn is the divergent harmonic series, 2::: a,.. is also divergent. n = l n.=l U I . n + 1 oo n! f"L-H>O an .,._.00 e n n. ft. - 00 en + n+ln! r&.-oc e n n= l en · ~ li I an+l I . 1' I (n + 1)! c" 2 1 1 . (n + 1)n! · e"2 1 25. se t 1e Ratto . est. m - - = un < +t) ~ · - 1 = rm 2 2 = 1m ~+t = 0 < 1, so :L ---:r converges. 27. roo ln: dx = lim [- ln X - .!.] t . [using integration by parts] M: 1. So E ln: converges by the Integral Test, and since ./ 2 X t-oo X X 1 n=l n k ln k k ln k ln k . . oo k In A: • -----,- 3 < - k 3 = -k?, the gtven senes 2::: 3 converges by the Companson Test. (k + l) .. ·- k=t(k+ l) 29. f a n = f (-1)"___!_h = f ( - 1)" b,,. Now b,. = - 1 - 1 - > 0, {b,..} is decreasing, and lim b, = 0, so the series n =l n=l cos n "·=1 cos 1 n n -oo converges by the Alternating Series Test. 0 W . 1 2 2 d ~ 1 '. . . ~ 1 . b h r: rtte co~ lt n = < - an 6 - ts a convergent geometrtc senes, so 6 -- 1 - ts convergent y t e ~ e" + e-" e" n = l e" n = l cos 1 n Comparison Test. So E ( n=l - 1)" ___!_h is absolutely convergent and therefore convergent. cos n 31. lim a~.; = lim k 5 4 k--+oo k-oo 3 + k k = (divide by 4") oo s" Thus, 2::: ~ diverges by the T
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.2 CALCULUS WITH PARAMETR
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SECTION 10.3 POLAR COORDINATES 0 13
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SECTION 10 .~ POLAR COORDINATES 0 1
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Page 39 and 40: x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
Page 41 and 42: SECTION 10.6 CONIC SECTIONS IN POLA
Page 43 and 44: CHAPTER 10 REVIEW 0 35 the length o
Page 45 and 46: CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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Page 51 and 52: 0 PROBLEMS PLUS l lt sin u dx cost
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Page 67 and 68: SECTION 11.3 THE INTEGRAL TEST AND
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Page 91 and 92: SECTION 11.10 TAYLOR AND MACLAURIN
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124 0 CHAPTER 12 VECTORS AND THE GE
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128 D CHAPTER 12 VECTORS AND THE GE
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136 D CHA~TER 12 VECTORS AND THE GE
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D PROBLEMS PLUS 1. Since three-dime
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l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
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13 D VECTOR FUNCTIONS 13.1 Vector F
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.2 DERIVATIVES AND INTEGR
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SECTION 13.2 DERIVATIVES AND INTEGR
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SECTION 13.3 ARC LENGTH AND CURVATU
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SECTION 13.4 MOTION IN SPACE: VELOC
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2 SECTION 13.4 MOTION IN SPACE: VEL
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CHAPTER 13 REVIEW D 173 43. The tan
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5. f~(t 2 i +tcos 1rtj +sin 1rt k )
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CHAPTER 13 REVIEW 0 177 23. (a) r (
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180 0 CHAPTER 13 PROBLEMS PLUS (-2a
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182 0 CHAPTER 13 PROBLEMS PLUS vect
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184 0 CHAPTER 14 PARTIAL DERIVATIVE
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196 D CHAPTER 14 PARTIAL DERIVATIVE
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D PROBLEMS PLUS 1. The areas of the
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15 D MULTIPLE INTEGRALS 15.1 Double
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SECTION 15.2 ITERATED INTEGRALS D 2
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l SECTION 15.3 DOUBLE INTEGRALS OVE
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SECTION 15.3 DOUBLE INTEGRALS OVER
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SECTION 15.3 DOUBLE INTEGRALS OVER
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61 . Since m :S j(x, y) :S M, ffD m
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SECTION 15.4 DOUBLE INTEGRALS IN PO
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SECTION 15.5 APPLICATIONS OF DOUBLE
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SECTION 15.5 APPLICATIONS OF DOUBLE
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SECTION 1505 APPLICATIONS OF DOUBLE
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-I SECTION 15.6 SURFACE AREA 0 267
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SECTION 15.7 TRIPLE INTEGRALS 0 269
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SECTION 15.7 TRIPLE INTEGRALS D 271
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Therefore E = { (x, y, z) I -2 ~ X~
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43. I,. = foL foL foL k(y2 + z2)dz.
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SECTION 15.8 TRIPLE INTEGRALS IN CY
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M xv = I~1f I: I:2 6 - 3 r 2 (zK) r
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SECTION 15.9 TRIPLE INTEGRALS IN SP
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SECTION 15.9 TRIPLE INTEGRALS IN SP
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(b) The wedge in question is the sh
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SECTION 15.10 CHANGE OF VARIABLES I
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CHAPTER 15 REVIEW 0 289 15 Review C
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CHAPTER 15 REVIEW 0 291 l 9. The vo
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CHAPTER 15 REVIEW D 293 33. Using t
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49. Since u = x- y and v = x + y, x
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298 D CHAPTER 15 PROBLEMS PLUS To e
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300 0 CHAPTER 15 PROBLEMS PLUS 13.
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16 0 VECTOR CALCULUS 16.1 Vector Fi
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SECTION 16.2 LINE INTEGRALS 0 305 2
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SECTION 16.2 LINE INTEGRALS 0 307 (
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SECTION 16.2 LINE INTEGRALS D 309 3
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SECTION 16.3 THE FUNDAMENTAL THEORE
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SECTION 16.4 GREEN'S THEOREM D 313
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SECTION 16.4 GREEN'S THEOREM 0 315
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8 8 8 (b)clivF = 'V ·iF=- (x+yz) +
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SECTION 16.5 CURL AND DIVERGENCE 0
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SECTION 16.6 PARAMETRIC SURFACES AN
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that is, D = {( x, y) I x 2 + y 2 :
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SECTION 16.7 SURFACE INTEGRALS 0 32
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SECTION 16.7 SURFACE INTEGRALS 0 33
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SECTION 16.8 STOKES' THEOREM 0 333
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dS SECTION 16.9 THE DIVERGENCE THEO
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CHAPTER 16 REVIEW 0 337 27. JI 5 cu
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CHAPTER 16 REVIEW 0 339 TRUE-FALSE
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CHAPTER 16 REVIEW D 341 Alternate s
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344 0 CHAPTER 16 PROBLEMS PLUS Simi
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3.c6 0 CHAPTER 17 SECOND-ORDER DIFF
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348 0 CHAPTER 17 SECOND-ORDER DIFFE
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352 D CHAPTER 17 SECOND-ORDER DIFFE
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356 D CHAPTER 17 SECOND-ORDER DIFFE
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0 APPENDIX Appendix H Complex Numbe
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APPENDIX H COMPLEX NUMBERS 0 361 43
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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