Solutions To Selected Problems

Chapter 9

Exercises 9.1

  1. 1.

    Answers will vary.

  2. 3.

    Answers will vary.

  3. 5.

    2,83,83,3215,6445

  4. 7.

    -13,-2,-815,-5123,-156257

  5. 9.

    an=3n+1

  6. 11.

    an=102n-1

  7. 13.

    1/7

  8. 15.

    0

  9. 17.

    diverges

  10. 19.

    converges to 0

  11. 21.

    converges to 0

  12. 23.

    diverges

  13. 25.

    converges to e

  14. 27.

    converges to 5

  15. 29.

    diverges

  16. 31.

    converges to 0

  17. 33.

    converges to 0

  18. 35.

    converges to ln2

  19. 37.

    converges to 0

  20. 39.

    bounded

  21. 41.

    bounded below

  22. 43.

    monotonically increasing

  23. 45.

    never monotonic

  24. 47.

    never monotonic

  25. 49.

    Let {an} be given such that limn|an|=0. By the definition of the limit of a sequence, given any ϵ>0, there is a m such that for all n>m,||an|-0|<ϵ. Since ||an|-0|=|an-0|, this directly implies that for all n>m, |an-0|<ϵ, meaning that limnan=0.

  26. 51.

    Left to reader

Exercises 9.2

  1. 1.

    Answers will vary.

  2. 3.

    One sequence is the sequence of terms {an}. The other is the sequence of nth partial sums, {Sn}={i=1nai}.

  3. 5.

    F

  4. 7.
    (a) -1,-12,-56,-712,-4760 (b) Plot omitted
  5. 9.
    (a) -1,0,-1,0,-1 (b) Plot omitted
  6. 11.
    (a) 1,32,53,4124,10360 (b) Plot omitted
  7. 13.
    (a) -0.9,-0.09,-0.819,-0.1629,-0.75339 (b) Plot omitted
  8. 15.

    Converges because it is a geometric series with r=15.

  9. 17.

    Diverges by Theorem 9.2.4

  10. 19.

    Diverges

  11. 21.

    limnan=1; by Theorem 9.2.4 the series diverges.

  12. 23.

    Diverges

  13. 25.

    Diverges

  14. 27.

    limnan=e; by Theorem 9.2.4 the series diverges.

  15. 29.

    Converges

  16. 31.

    Converges

  17. 33.
    (a) Sn=1-(1/4)n3/4 (b) Converges to 4/3.
  18. 35.
    (a) Sn={-n+12n is odd n2n is even (b) Diverges
  19. 37.
    (a) Sn=1-(1/e)n+11-1/e. (b) Converges to 1/(1-1/e)=e/(e-1).
  20. 39.
    (a) With partial fractions, an=1n-1n+1. Thus Sn=1-1n+1. (b) Converges to 1.
  21. 41.
    (a) Use partial fraction decomposition to recognize the telescoping series: an=12(12n-1-12n+1), so that Sn=12(1-12n+1)=n2n+1. (b) Converges to 1/2.
  22. 43.
    (a) Sn=1-1(n+1)2 (b) Converges to 1.
  23. 45.
    (a) an=1/2n+1/3n for n0. Thus Sn=1-1/221/2+1-1/3n2/3. (b) Converges to 2+3/2=7/2.
  24. 47.
    (a) Sn=1-(sin1)n+11-sin1 (b) Converges to 11-sin1.
  25. 49.

    (-3,3)

  26. 51.

    (-,-4)(4,)

  27. 53.

    Using partial fractions, we can show that an=14(12n-1+12n+1). The series is effectively twice the sum of the odd terms of the Harmonic Series which was shown to diverge in Example 9.2.5. Thus this series diverges.

Exercises 9.3

  1. 1.

    continuous, positive and decreasing

  2. 3.

    Converges

  3. 5.

    Diverges

  4. 7.

    Converges

  5. 9.

    Converges

  6. 11.

    p>1

  7. 13.

    p>1

Exercises 9.4

  1. 1.

    n=0bn converges; we cannot conclude anything about n=0cn

  2. 3.

    Converges; compare to n=11n2, as 1/(n2+3n-5)1/n2 for all n>1.

  3. 5.

    Diverges; compare to n=11n, as 1/nlnn/n for all n3.

  4. 7.

    Diverges; compare to n=11n. Since n=n2>n2-1, 1/n1/n2-1 for all n2.

  5. 9.

    Converges; compare to n=11n2.

  6. 11.

    Diverges; compare to n=1lnnn.

  7. 13.

    Diverges; compare to n=11n.

  8. 15.
    Diverges; compare to n=11n: 1n=n2n3<n2+n+1n3<n2+n+1n3-5, for all n1.
  9. 17.

    Converges; compare to n=1(25)n, as 2n/(5n+10)<2n/5n for all n1.

  10. 19.

    Converges by Comparison Test with 1n3

  11. 21.
    Diverges; compare to n=11n. Note that nn2-1=n2n2-11n>1n, as n2n2-1>1, for all n2.
  12. 23.

    Converges; compare to n=11n2, as 1/(n2lnn)1/n2 for all n3.

  13. 25.

    Converges; Integral Test

  14. 27.

    Diverges; the nth Term Test and Direct Comparison Test can be used.

  15. 29.

    Converges; the Direct Comparison Test can be used with sequence 1/3n.

  16. 31.

    Diverges; the nth Term Test can be used, along with the Integral Test.

  17. 33.
    (a) Converges; use Direct Comparison Test as ann<an. (b) Converges; since original series converges, we know limnan=0. Thus for large n, anan+1<an. (c) Converges; similar logic to part (b) so (an)2<an. (d) May converge; certainly nan>an but that does not mean it does not converge. (e) Does not converge, using logic from (b) and nth Term Test.

Exercises 9.5

  1. 1.

    The signs of the terms do not alternate; in the given series, some terms are negative and the others positive, but they do not necessarily alternate.

  2. 3.

    Many examples exist; one common example is an=(-1)n/n.

  3. 5.
    (a) converges (b) converges (p-Series) (c) absolute
  4. 7.
    (a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges
  5. 9.
    (a) converges (b) diverges (Limit Comparison Test with 1/n) (c) conditional
  6. 11.
    (a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges
  7. 13.
    (a) diverges (terms oscillate between ±1) (b) diverges (c) n/a; diverges
  8. 15.
    (a) converges (b) converges (Geometric Series with r=2/3) (c) absolute
  9. 17.
    (a) converges (b) diverges (p-Series Test with p=1/2) (c) conditional
  10. 19.

    S5=-1.1906; S6=-0.6767;

    -1.1906n=1(-1)nln(n+1)-0.6767

  11. 21.

    S6=0.3681; S7=0.3679;

    0.3681n=0(-1)nn!0.3679

  12. 23.

    n=5

  13. 25.

    Using the theorem, we find n=499 guarantees the sum is within 0.001 of π/4. (Convergence is actually faster, as the sum is within ϵ of π/24 when n249.)

Exercises 9.6

  1. 1.

    algebraic, or polynomial.

  2. 3.

    Integral Test, Limit Comparison Test, and Root Test

  3. 5.

    Converges

  4. 7.

    Converges

  5. 9.

    The Ratio Test is inconclusive; the p-Series Test states it diverges.

  6. 11.

    Converges

  7. 13.

    Converges; note the summation can be rewritten as n=12nn!3nn!, from which the Ratio Test can be applied.

  8. 15.

    Diverges

  9. 17.

    Converges

  10. 19.

    Converges

  11. 21.

    Diverges

  12. 23.

    Diverges. The Root Test is inconclusive, but the nth-Term Test shows divergence. (The terms of the sequence approach e-2, not 0, as n.)

  13. 25.

    Converges

Exercises 9.7

  1. 1.

    Diverges

  2. 3.

    Diverges

  3. 5.

    Diverges

  4. 7.

    Absolutely converges

  5. 9.

    Conditionally converges

  6. 11.

    Diverges

  7. 13.

    Absolutely converges

  8. 15.

    Absolutely converges

  9. 17.

    Absolutely converges

  10. 19.

    Conditionally converges

  11. 21.

    Absolutely converges

  12. 23.

    Absolutely converges

  13. 25.

    Diverges

  14. 27.

    Diverges

  15. 29.

    Absolutely converges

  16. 31.

    Diverges

  17. 33.

    Absolutely converges

  18. 35.

    Diverges

  19. 37.

    Absolutely converges

Exercises 9.8

  1. 1.

    1

  2. 3.

    5

  3. 5.

    1+2x+4x2+8x3+16x4

  4. 7.

    1+x+x22+x36+x424

  5. 9.
    (a) R= (b) (-,)
  6. 11.
    (a) R=1 (b) (2,4]
  7. 13.
    (a) R=2 (b) (-2,2)
  8. 15.
    (a) R=1/5 (b) (4/5,6/5)
  9. 17.
    (a) R=1 (b) (-1,1)
  10. 19.
    (a) R= (b) (-,)
  11. 21.
    (a) R=1 (b) [-1,1]
  12. 23.
    (a) R=0 (b) x=0
  13. 25.
    (a) R=1 (b) [-13,53)
  14. 27.
    (a) R= (b) (-,)
  15. 29.

    n=08nxn+1, R=1/8

  16. 31.

    n=0x2n+33n+1, R=3

  17. 33.
    (a) n=0xn(-1)n(1+n); R=1 (b) n=1xn-1(-1)n-1(1+n)n/2=n=0xn(-1)n(2+n)(1+n)/2 (c) n=1xn+1(-1)n-1(1+n)n/2=n=0xn+2(-1)n(2+n)(1+n)/2=n=2xn(-1)nn(-1+n)/2
  18. 35.

    n=1n(-9)n-1xn, R=1/9

  19. 37.

    n=0(-1)nx2n+12n+1; R=1

  20. 39.

    n=0(2n+1)xn; R=1

Exercises 9.9

  1. 1.

    The Maclaurin polynomial is a special case of Taylor polynomials. Taylor polynomials are centered at a specific x-value; when that x-value is 0, it is a Maclaurin polynomial.

  2. 3.

    p2(x)=6+3x-4x2.

  3. 5.

    p3(x)=1-x+12x3-16x3

  4. 7.

    p8(x)=x+x2+12x3+16x4+124x5

  5. 9.

    p4(x)=2x43+4x33+2x2+2x+1

  6. 11.

    p4(x)=x4-x3+x2-x+1

  7. 13.

    p4(x)=1+12(-1+x)-18(-1+x)2+116(-1+x)3-5128(-1+x)4

  8. 15.

    p6(x)=12--π4+x2-(-π4+x)222+(-π4+x)362+(-π4+x)4242-(-π4+x)51202-(-π4+x)67202

  9. 17.

    p5(x)=12-x-24+18(x-2)2-116(x-2)3+132(x-2)4-164(x-2)5

  10. 19.

    p3(x)=12+1+x2+14(1+x)2

  11. 21.

    p3(x)=x-x36; p3(0.1)=0.09983. Error is bounded by 14!0.140.000004167.

  12. 23.

    p2(x)=3+16(-9+x)-1216(-9+x)2; p2(10)=3.16204. The third derivative of f(x)=x is bounded on [9,10] by 0.0015. Error is bounded by 0.00153!13=0.0003.

  13. 25.

    The nth derivative of f(x)=ex is bounded by e on [0,1]. Thus |Rn(1)|e(n+1)!1(n+1). When n=7, this is less than 0.0001.

  14. 27.

    The nth derivative of f(x)=cosx is bounded by 1 on intervals containing 0 and π/3. Thus |Rn(π/3)|1(n+1)!(π/3)(n+1). When n=7, this is less than 0.0001. Since the Maclaurin polynomial of cosx only uses even powers, we can actually just use n=6.

  15. 29.

    1n!xn

  16. 31.

    When n even, 0; when n is odd, (-1)(n-1)/2n!xn.

  17. 33.

    (-1)nxn

  18. 35.

    1+x+12x2+16x3+124x4

Exercises 9.10

  1. 1.

    A Taylor polynomial is a polynomial, containing a finite number of terms. A Taylor series is a series, the summation of an infinite number of terms.

  2. 3.
    All derivatives of ex are ex which evaluate to 1 at x=0. The Taylor series starts 1+x+12x2+13!x3+14!x4+; the Taylor series is n=0xnn!
  3. 5.
    The nth derivative of 1/(1-x) is f(n)(x)=(n)!/(1-x)n+1, which evaluates to n! at x=0. The Taylor series starts 1+x+x2+x3+; the Taylor series is n=0xn
  4. 7.
    The Taylor series starts 0-(x-π/2)+0x2+16(x-π/2)3+0x4-1120(x-π/2)5; the Taylor series is n=0(-1)n+1(x-π/2)2n+1(2n+1)!
  5. 9.
    f(n)(x)=(-1)ne-x; at x=0, f(n)(0)=-1 when n is odd and f(n)(0)=1 when n is even. The Taylor series starts 1-x+12x2-13!x3+; the Taylor series is n=0(-1)nxnn!.
  6. 11.
    f(n)(x)=(-1)n+1n!(x+1)n+1; at x=1, f(n)(1)=(-1)n+1n!2n+1 The Taylor series starts 12+14(x-1)-18(x-1)2+116(x-1)3; the Taylor series is 12+n=1(-1)n+1(x-1)n2n+1.
  7. 13.
    Given a value x, the magnitude of the error term Rn(x) is bounded by |Rn(x)|max|f(n+1)(z)|(n+1)!|x(n+1)|, where z is between 0 and x.

    If x>0, then z<x and f(n+1)(z)=ez<ex. If x<0, then x<z<0 and f(n+1)(z)=ez<1. So given a fixed x value, let M=max{ex,1}; f(n)(z)<M. This allows us to state

    |Rn(x)|M(n+1)!|x(n+1)|.

    For any x, limnM(n+1)!|x(n+1)|=0. Thus by the Squeeze Theorem, we conclude that limnRn(x)=0 for all x, and hence

    ex=n=0xnn!for all x.
  8. 15.
    Given a value x, the magnitude of the error term Rn(x) is bounded by |Rn(x)|max|f(n+1)(z)|(n+1)!|x(n+1)|, where z is between 0 and x. Since |f(n+1)(z)|=n!(z+1)n+1, |Rn(x)|1n+1(|x|minz+1)n+1.

    If 0<x<1, then 0<z<x and f(n+1)(z)=n!(z+1)n+1<n!. Thus

    |Rn(x)|n!(n+1)!|x(n+1)|=xn+1n+1.

    For a fixed x<1,

    limnxn+1n+1=0.
  9. 17.
    Given cosx=n=0(-1)nx2n(2n)!, cos(-x)=n=0(-1)n(-x)2n(2n)!=n=0(-1)nx2n(2n)!=cosx, as all powers in the series are even.
  10. 19.
    Given sinx=n=0(-1)nx2n+1(2n+1)!, ddx(sinx)=ddx(n=0(-1)nx2n+1(2n+1)!)=n=0(-1)n(2n+1)x2n(2n+1)!=n=0(-1)nx2n(2n)!=cosx. (The summation still starts at n=0 as there was no constant term in the expansion of sinx).
  11. 21.

    1+x2-x28+x316-5x4128

  12. 23.

    1+x3-x29+5x381-10x4243

  13. 25.

    n=0(-1)n(x2)2n(2n)!=n=0(-1)nx4n(2n)!.

  14. 27.

    n=0(-1)n(2x+3)2n+1(2n+1)!.

  15. 29.

    x+x2+x33-x530

  16. 31.

    0πsin(x2)𝑑x0π(x2-x66+x10120-x145040)𝑑x=0.8877

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