Sequências Infinitas e Séries resolvidas - Thomas

Sequências Infinitas e Séries resolvidas - Thomas

(Parte 1 de 3)

CHAPTER 1 INFINITE SEQUENCES AND SERIES

1.1 SEQUENCES

15. a(1)n, n1, 2, 16. a, n1, 2, n n1 () n

23. lim 2(0.1)2 converges (Theorem 5, #4) nÄ_ œ Ên

698 Chapter 1 Infinite Sequences and Series

24. lim lim 1 converges nnÄ_ Ä_ n( ) ( 1)n

25. lim lim lim 1 converges n nÄ_ Ä_ Ä_

26. lim limdiverges

È n n

27. lim lim 5 converges nnÄ_ Ä_"

28. lim lim lim 0 converges n nÄ_ Ä_ Ä_

29. lim lim lim (n1) diverges n nÄ_ Ä_ Ä_

30 lim limdiverges

4 œœ _ ÊŠ‹Š‹

3. lim 1 lim 1 converges nnÄ_ Ä_

34. lim 236 converges 35. lim 0 converges nnÄ_ Ä_

36. lim lim 0 converges

37. lim lim lim 2 converges

É É Ê Š‹ È2n 2n

38. lim lim diverges

39. lim sinsin lim sin 1 converges nnÄ_ Ä_

41. lim 0 because converges by the Sandwich Theorem for sequences nÄ_ sin nsin n

42. lim 0 because 0 converges by the Sandwich Theorem for sequences nÄ_ sin n sin n ## #"œŸ Ÿ Ê

43. lim lim 0 converges (using l'Hopital's rule)^

4. lim lim lim limdiverges (using l'Hopital's rule)^

45. lim lim lim lim 0 converges

" n1 n È

Section 1.1 Sequences699

46. lim lim 1 converges nnÄ_ Ä_ ln n

47. lim 81 converges (Theorem 5, #3) nÄ_ 1nÎ œÊ

48. lim (0.03)1 converges (Theorem 5, #3) nÄ_ 1nÎ œÊ

49. lim 1e converges (Theorem 5, #5)

50. lim 1 lim 1e converges (Theorem 5, #5)

51. lim 10n lim 10n111 converges (Theorem 5, #3 and #2) nnÄ_ Ä_

È œœ œ Ê1n 1nÎÎ ††

52. lim n lim n11 converges (Theorem 5, #2) nnÄ_ Ä_ È ‰È# # #œœ œ Ê

53. lim 1 converges (Theorem 5, #3 and #2)

1n lim 3 lim n Î "œœ œ Ênn

54. lim (n4) lim x1 converges; (let xn4, then use Theorem 5, #2) n xÄ_ Ä_ œ œ Ê œ 1n 4 1 xÎÐ Ñ Î

5. limdiverges (Theorem 5, #2)

ln n n lim n lim ln n

1 œœ œ_ Ênn

56. lim ln nln(n1) lim lnln lim ln 10 converges n nÄ_ Ä_ Ä_

57. lim 4n lim 4n414 converges (Theorem 5, #2) nnÄ_ Ä_ È Èn œœ œ ʆ

59. lim lim lim 0 and 0 lim 0 converges n n nÄ_ Ä_ Ä_ Ä_ n! n! n! n n n n n n 2 3 (n 1)(n)œŸ œ Ê œ Ê"â â

60. lim 0 converges (Theorem 5, #6)

61. lim limdiverges (Theorem 5, #6)
62. lim limdiverges (Theorem 5, #6)

63. lim lim exp ln lim expe converges n nÄ_ Ä_ Ä_ nln nnln n 1lnnln 1ln nœœœÊ

64. lim ln1ln lim 1ln e1 converges (Theorem 5, #5)

65. lim lim expn ln lim exp n nÄ_ Ä_ Ä_

700 Chapter 1 Infinite Sequences and Series lim exp lim expexpe convergesœœœœÊ nnÄ_ Ä_ Š‹ ‰3

# Š‹ 6n 6

6. lim lim expn ln lim exp lim exp lim expe convergesœ œÊ nÄ_

67. lim lim x lim exp lnx lim exp n n nÄ_ Ä_ Ä_ Ä_

x lim expxex, x0 convergesœœœ Ê

68. lim 1 lim expn ln1 lim exp lim exp n n lim expe1 convergesœœœÊ

!2n n1

69. lim lim 0 converges (Theorem 5, #6) nnÄ_ Ä_

70. lim lim lim 0 converg

(Theorem 5, #4)

71. lim tanh n lim lim lim lim 1 converges n n n nÄ_ Ä_ Ä_ Ä_ Ä_ œœ œ œ " œ Êee e 2e

72. lim sinh(ln n) lim limdiverges
73. lim lim lim limconverges

n sin sin coscos‰ ˆ‰ n n

74. lim n1cos lim lim lim sin0 converges n n nÄ_ Ä_ Ä_ Ä_ n n

75. lim tann converges 76. lim tann00 converges nnÄ_ Ä_

7. lim lim 0 converges (Theorem 5, #4)

78. lim n lim exp lim expe1 converges

2n 1ab

79. lim lim lim lim 0 converges n n nÄ_ Ä_ Ä_ Ä_

(ln n)200(ln n)200199(ln n) n n n

200! œœ œá œ œ ʆ

80. lim lim lim lim lim 0 converges n n n nÄ_ Ä_ Ä_ Ä_ Ä_

(ln n)10(ln n)80(ln n) n n n 3840ÈÈ È Èœ œœœ á œ œ Ê–—Š‹Š‹

Section 1.1 Sequences701 n n n n n n n 1È É "

convergesœÊ" #

82. lim lim lim n nÄ_ Ä_ Ä_"" È È È È n 1 n n 1 n n1 n n 1 n n1 n 1nœœ Š‹ Š ‹

lim 2 convergesœœ Ê

83. limdx lim lim 0 converges (Theorem 5, #1)

nx n n ln n'1 n œœ œ Ê n n

a2 bab

In the first and second fractions, yn. Let represent the (n1)th fraction where 1 and bn1n a b for n a positive integer3. Now the nth fraction is and ab2b2n2n yn. Thus, Ê a2bab n lim r2. nÄ_ n œ È

(b) f(x)tan(x)1; the sequence converges to 0.7853981635œ 1 4

8. (a) lim nf lim lim f(0), where x n xxÄ_ Ä! Ä!

(b) lim n tanf(0)1, f(x)tanx

(c) lim ne1f(0)e1, f(x)e1 nÄ_ ab1n xÎ œ œ œ œ w!

(d) lim n ln1f(0)2, f(x)ln(12x) nÄ_

(b) lim lim 1 or lim lim sin lim sin 1 a a 2Ä_ Ä_ Ä_ Ä_ ÄÎ

90. (a) lim (2n) lim exp lim exp lim exp e1; n n nÄ_ Ä_ Ä_ Ä_ n! 2n, Stirlings approximation n!(2n) for large values of nʉ‰Èn e e 12 n11 Îab

(b)n n!

È n e

702 Chapter 1 Infinite Sequences and Series

91. (a) lim lim lim 0 n nÄ_ Ä_ Ä_ ln n nc n cnœœ œ

(b) For all 0, there exists an Ne such that ne ln n ln nln% œ Ê Ê ÐÑÎ ÐÑÎ"lnclnccln c %% % %

n0 lim 0n

n n

n1, 2, 3,For all 0 there exists N such that when nN then aL and there exists Nχ %%""#kkn

92. Let {a} and {b} be sequences both converging to L. Define {c} by cb and ca, wherennn2nn2n1nœœ such that when nN then bL. If n12max{N}, then cL, so {c} converges to L. ß #"#kn%%

93. lim n lim exp ln n lim expe1 n nÄ_ Ä_ Ä_1n n Î! ""œœ œ œ‰ ‰

94. lim x lim exp ln xe1, because x remains fixed while n gets large nnÄ_ Ä_1n n Î! "œœ œˆ‰

96. Let . We have f continuous at Lthere exists so that xLf(x)f(L). Also, aLthere%$$% !Ê Ê ÄÊkkkkn exists N so that for nN aL. Thus for nN, f(a)f(L) f(a) f(L). ÊÄkkkknnn$%

97. a3n3n4n43n6nn2n1n

42; the steps are reversible so the sequence is nondecreasing; 3 3n13n3Ê Ê 3nn1 " 13; the steps are reversible so the sequence is bounded above by 3Ê

98. an1n
9. a23n1 which is true for n5; the steps aren1n

(n 1)! reversible so the sequence is decreasing after a, but it is not nondecreasing for all its terms; a6, a18,&"#œœ a36, a54, a64.8 the sequence is bounded from above by 64.8$%&œœœœÊ324 5

100. a 2; the steps aren1n

101. a1 converges because 0 by Example 1; also it is a nondecreasing sequence bounded above by 1nœ Ä"" n

102. an diverges because n and 0 by Example 1, so the sequence is unboundednœ Ä_Ä"" n

103. a1 and 0; since 0 (by Example 1)0, the sequence converges; also it isn21

2n n a nondecreasing sequence bounded above by 1

Section 1.1 Sequences703

diverges by definition of divergence

106. xmax{cos 1cos 2cos 3cos n} and xmax{cos 1cos 2cos 3cos(n1)}x with x1nn1nnœßßßáßœßßßáß Ÿ so the sequence is nondecreasing and bounded above by 1 the sequence converges.Ê

107. If {a} is nonincreasing with lower bound M, then {a} is a nondecreasing sequence with upper bound M.n

By Theorem 1, {a} converges and hence {a} converges. If {a} has no lower bound, then {a} has no n upper bound and therefore diverges. Hence, {a} also diverges.n

108. an2n1n2n 10 and 1; thus the sequence isnn1

n1 n1 nonincreasing and bounded below by 1 it convergesÊ

109. an12n2nn2n2n n1nnn1

and 2; thus the sequence is nonincreasing and bounded below by 2 it converges12nn ÈÈ ÊÈÈ

110. a224224 222424nn1
1. 4 so a 41 and43333333

so the sequence is nonincreasing but not bounded below and therefore diverges

113. Let 0M1 and let N be an integer greater thanThen nN n nM Ê Ê M

114. Since M is a least upper bound and M is an upper bound, M. Since M is a least upper bound and M"#"##"Ÿ is an upper bound, M. We conclude that M so the least upper bound is unique.#""#Ÿœ

115. The sequence a1 is the sequence , , , ,This sequence is bounded above by ,n

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

116. Let L be the limit of the convergent sequence {a}. Then by definition of convergence, for theren% #

704 Chapter 1 Infinite Sequences and Series

118. 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 and a, where a L,kninknÐÑÐÑÐÑ"Ä a L and L. Thus aaL0. So there does not exist N such that for all m, nNinkninÐÑÐÑÐÑ#"#"#ÄÁ Ä k a. So by Exercise 116, the sequence a is not convergent and hence diverges.Ê Ö×kkmnn%

119. a L given an 0 there corresponds an N such that 2kN aL. Similarly,2k2kÄÍ Ê %%""cdkk a L 2k1N aL. Let Nmax{N}. Then nN aL whether2k12k1n # "#ÄÍ Ê œß Ê cdkkkk%% n is even or odd, and hence a L.nÄ

120. Assume a 0. This implies that given an 0 there corresponds an N such that nN a0nnÄ Ê %%k a a a0 a 0. On the other hand, assume a 0. This implies thatÊ Ê Ê ÊÄÄkkkkkkkkkkkkkknnnnn%%% given an 0 there corresponds an N such that for nN, a0 a a%%%% Ê Ê kn a0 a 0.Ê ÊÄkknn%

121. 0.5110 1n n692.8¹¹È‰‰ˆ‰

1n 9 1001nn lnln ‰ˆ‰

N692; a and lim a1Êœœœnn1n‰"#Î nÄ_

1n 9 1001nn

ann and lim a1nn1nœœœÈÎ nÄ_

Ê  ÊÊ œ œ œ$  ˆ‰
Ê œ œ œ( (

124. 10 n!210 and by calculator experimentation, n14 N14; a and lim a022 n! n!n n nÄ_ xa 2x xa xa x2 x 2x xab ˆ‰

126. x1.5, x1.416666667, x1.414215686, x1.414213562, x1.414213562; we are finding the"#$%&œœœœœ positive number x20; that is, where x2, x0, or where x2.## œœ œÈ x1.570791601cos(1.570791601)1.570796327 to 9 decimal places. After a few steps, the%#œ œœ1 arcx and line segment cosx are nearly the same as the quarter circle.ababn1n1

129-140. Example CAS Commands: :Maple with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi;

Section 1.1 Sequences705

N := [ 100, 200, 1000 ];# (b)
for n in N do# (c)
FunctionAverage(f(x),x=a..b,output=plot);# (d)

plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); : (sequence functions may vary):Mathematica

Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n,1,25}] Limit[a[n], n8]Ä

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, do the following.

Clear[minN, lim] lim= 1

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.

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

A(0) := 1000; r := 0.02015; m := 12; b := 50;# (a)

141. Example CAS Commands: :Maple with( Student[Calculus1] ); A := n->(1+r/m)*A(n-1) + b; A(0) := A0; pts1 := [seq( [n,A(n)], n=0..9 )]: plot( pts1, style=point, title="#141(a) (Section 1.1)");

706 Chapter 1 Infinite Sequences and Series

A(60);

A(0) := 5000; r := 0.0589; m := 12; b := -50;# (b)
A(0) := 5000; r := 0.045; m := 4; b := 0;# (c)

The sequence { A[n] } is not unbounded; limit( A[n], n=infinity ) = infinity. pts1 := [seq( [n,A(n)], n=0..9 )]: plot( pts1, style=point, title="#141(b) (Section 1.1)"); A(60); pts1 := [seq( [n,A(n)], n=0..199 )]: plot( pts1, style=point, title="#141(b) (Section 1.1)"); # This sequence is not bounded, and diverges to -infinity: limit( A[n], n=infinity ) = -infinity. for n from 1 while A(n)<20000 do end do; n;

It takes 31 years (124 quarters) for the investment to grow to $20,0 when the interest rate is 4.5%, compounded quarterly. r := 0.0625; for n from 1 while A(n)<20000 do end do; n;

B := k -> (1+r/m)k * (A(0)+m*b/r) - m*b/r;# (d)

When the interest rate increases to 6.25% (compounded quarterly), it takes only 2.5 years for the balance to reach $20,0. A(0) := 1000.; r := 0.02015; m := 12; b := 50; for k from 0 to 49 do printf( "%5d %9.2f %9.2f %9.2f\n", k, A(k), B(k), B(k)-A(k) ); end do;

A(0) := 'A(0)'; r := 'r'; m := 'm'; b := 'b'; n := 'n'; eval( A(n+1) - ((1+r/m)*A(n) + b), A=B ); simplify( % );

r := 3/4.;# (a)
A := r*A*(1-A);
L := L, [n,A];
R1 := [1.1, 1.2, 1.5, 2.5, 2.8, 2.9];# (b)
A := k/10.;
L := [0,A];
for n from 1 to 9 do
A := r*A*(1-A);

142. Example CAS Commands: :Maple for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 9 do end do; pt[r,k/10] := [L]; end do: plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title="#142(a) (Section 1.1)" ); for r in R1 do for k in $1..9 do L := L, [n,A];

end do;
pt[r,k/10] := [L];
R2 := [3.05, 3.1, 3.2, 3.3, 3.35, 3.4];# (c)
A := k/10.;
L := [0,A];
for n from 1 to 9 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
R3 := [3.46, 3.47, 3.48, 3.49, 3.5, 3.51, 3.52, 3.53, 3.542, 3.544, 3.546, 3.548];# (d)
A := k/10.;
L := [0,A];
for n from 1 to 199 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
R4 := [3.5695];# (e)
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];

Section 1.1 Sequences707 end do: t := sprintf("#142(b) (Section 1.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R1)], insequence=true ); for r in R2 do for k in $1..9 do end do: t := sprintf("#142(c) (Section 1.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R2)], insequence=true ); for r in R3 do for k in $1..9 do end do: t := sprintf("#142(d) (Section 1.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R3)], insequence=true ); for r in R4 do for k in $1..9 do end do: t := sprintf("#142(e) (Section 1.1)\nr = %f", r);

708 Chapter 1 Infinite Sequences and Series

R5 := [3.65];# (f)
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
R6 := [3.65, 3.75];# (g)
A := a;
L := [0,a];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,a] := [L];

P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R4)], insequence=true ); for r in R5 do for k in $1..9 do end do: t := sprintf("#142(f) (Section 1.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R5)], insequence=true ); for r in R6 do for a in [0.300, 0.301, 0.600, 0.601 ] do end do: t := sprintf("#142(g) (Section 1.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[0.300, 0.301, 0.600, 0.601] )], style=point, title=t ); end do: display( [seq(P[r], r=R6)], insequence=true );

1.2 INFINITE SERIES

1. s lim s3nn a1 r

(1 r)

2. s lim snn a1 r

(1 r)1

3. s lim snn a1 r

5s lim s""""""""""""
6s555555555555555

lim s5Êœ nÄ_ n

15s14
16A(2n1)B(2n1)66AB

(2n 1)(2n 1)2n 12n 1(2n 1)(2n 1)

(2A2B)n(AB)62A6 A3 and B3. Hence,2A2B0AB0

3 3!!‰Š‹k n1 n1œœ

œA(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1) (2n 1)(2n 1)

8A8C 08A8C 0

4A4B4C4D 0 A BC D 0 2A 4B 2C 4D 40 A 2 A B C D 0 œ ÚÝÛÝÜ BC2 D 20 2B2 D 20 A BC D 0

and D5C0 and A0. Hence,

œœ ! ’“k n1œ

710 Chapter 1 Infinite Sequences and Series

18s12n 1

24. divergent geometric series with r21 25. convergent geometric series with sum 1kkÈœ œŠ‹Š‹

29. convergent geometric series with sum "

1 ee1Š‹" #e

30. lim ln 0 diverges n Ä _ "n œ _Á Ê

32. convergent geometric series with sum "

1 x x 1Š‹"

3. difference of two geometric series with sum 3"" ##1 1 3Š‹Š‹23

34. lim 1 lim 1e0 diverges

Section 1.2 Infinite Series71

35. lim 0 diverges 36. lim lim lim n diverges n n nÄ_ Ä_ Ä_ Ä_ n! n n n

38. lim a lim lnln0 diverges

39. convergent geometric series with sum "

1e‰e1
42(1)x x; a1, rx; converges to for x1!!abkk__

œœn0 n0

4; a, r; converges to !!‰‰_

œœn0 n0

3 sin x for all x since for all xœœŸŸ3 sin x3 sin x 2(4 sin x)8 2 sin x43 sin x

45. a1, r2x; converges to for 2x1 or xœœ "" #1 2xkkkk

49. a1, r sin x; converges to for x(2k1), k an integerœœÁ " #1 sin x

51. 0.23 52. 0.234 œœœœœœ!!‰‰_ œ œn0 n0 n n

53. 0.7 54. 0.d œœœœœœ!!‰‰_ œ œn0 n0 n n

712 Chapter 1 Infinite Sequences and Series

œn0

59. (a) (b) (c) !!!_

"" " #(n 4)(n 5) (n 2)(n 3) (n 3)(n )

60. (a) (b) (c) !!!_

61. (a) one example is 1""""

negative number.

62. The series k is a geometric series whose sum is k where k can be any positive or negative number.!‰_

63. Let ab. Then a b 1, while (1) diverges.n

œœ œ œ œn1 n1 n1 n1 n1

64. Let ab. Then a b 1, while ab AB.n n n

œœ œ œ œn1 n1 n1 n1 n1

œœ œ œn1 n1 n1 n1

6. Yes:diverges. The reasoning: a converges a 0 diverges by the!!!Š‹Š‹"""

a an Ê Ä Ê Ä_Ê nth-Term Test.

67. 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.

contradicts the assumption that b diverges; therefore, ab diverges.!!abnnn

Section 1.3 The Integral Test713

(b) 5 1r r;

1r 1r 1r a1 r ab

AA34, AA34,,%$""œœababŠ‹Š‹‰‰2333

78. Each term of the series represents the area of one of the squares shown in the figure, and all of the!_ œn1 œ n0

1.3 THE INTEGRAL TEST

714 Chapter 1 Infinite Sequences and Series

3. diverges; by the nth-Term Test for Divergence, lim 10

4. diverges by the Integral Test; dx5 ln(n1)5 ln 2 dx ''1

œœn1 n13nnÈÈœœ "" #

6. converges; 2 , which is a convergent p-series (p)!!_ œœn1 n1

7. converges; a geometric series with r1œ " 8

8. diverges; 8and since diverges, 8 diverges!!!!_

œœ œ œn1 n1 n1 n1

9. diverges by the Integral Test: dxlnnln 2 dx

10. diverges by the Integral Test: dx; te dt lim 2te4e tln x dt dxe dt

2l n2 __ln xxdx

1. converges; a geometric series with r1œ 2 3

12. diverges; lim lim lim 0 n nÄ_ Ä_ Ä_

5 ln 5ln 5

13. diverges; 2 , which diverges by the Integral Test!!_ œœn0 n0

14. diverges by the Integral Test: ln(2n1) as n '1

15. diverges; lim a lim lim 0 n nÄ_ Ä_ Ä_n

2 ln 2 n1 1œœ œ _ Á

16. diverges by the Integral Test: ; lnn1ln 2uxdu''12 n 1dx du as n Ä_Ä_

17. diverges; lim lim lim 0 n nÄ_ Ä_ Ä_ È Š‹Š‹n

ln n

18. diverges; lim a lim 1e0 nnÄ_ Ä_n n

20. converges; a geometric series with r0.911œ " ln 3

21. converges by the Integral Test: dx;du

Section 1.3 The Integral Test715 uln x du dx'' 3l n3 _ Š‹È"

(ln x)(ln x)1x

lim sec u lim secbsec(ln 3) lim cossec(ln 3)œœ œ b bÄ_ Ä_ Ä_

2. converges by the Integral Test: dx dx;du

uln x du dx ab Š‹ œÄ

23. diverges by the nth-Term Test for divergence; lim n sin lim lim 10 sin xœœœÁ

24. diverges by the nth-Term Test for divergence; lim n tan lim lim œœ tan sec‰ˆ‰ n n lim secsec010œœœÁ nÄ_

25. converges by the Integral Test: dx;du lim tanuue

due dx''1ex x œ Äœ nÄ_ b e

26. converges by the Integral Test: dx; du duue due dx dx du

(Parte 1 de 3)

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