Section 11.3 The Integral Test and Estimates of Sums

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Section 11.3 The Integral Test and Estimates of Sums

29. Explain why the Integral Test cant be used to determine whether the series is convergent.

∞

X

n=1

cos πn

√n

Solution:

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ¤ 133 26. The function () = 

⁴+ 1 is positive, continuous, and decreasing on [1 ∞). [Note that

⁰() = ⁴+ 1 − 4⁴

(⁴+ 1)² = 1 − 3⁴

(⁴+ 1)²  0on [1 ∞).] Thus, we can apply the Integral Test.

 ∞ 1



⁴+ 1 = lim

→∞

  1

1 2(2)

1 + (²)²  = lim

→∞

1

2tan⁻¹(²)

 1

= 1 2 lim

→∞[tan⁻¹(²) − tan⁻¹1] = 1 2

  2 −

4



=  8 so the series ^∞

=1



⁴+ 1 converges.

27. The function () = cos 

√ is neither positive nor decreasing on [1 ∞), so the hypotheses of the Integral Test are not

satisfied for the series ^∞

=1

cos 

√ .

28. The function () = cos²

1 + ² is not decreasing on [1 ∞), so the hypotheses of the Integral Test are not satisfied for the series ^∞

=1

cos² 1 + ².

29. We have already shown (in Exercise 21) that when  = 1 the series ^∞

=2

1

(ln )^ diverges, so assume that  6= 1.

 () = 1

(ln )^ is continuous and positive on [2 ∞), and ⁰() = −  + ln 

²(ln )^+1  0if   ^−, so that  is eventually decreasing and we can use the Integral Test.

 ∞ 2

1

(ln )^  = lim

→∞

(ln )¹^−

1 − 

^

2

[for  6= 1] = lim_

→∞

(ln )¹^−

1 −  −(ln 2)¹^−

1 − 



This limit exists whenever 1 −   0 ⇔   1, so the series converges for   1.

30. () = 1

 ln  [ln(ln )]^ is positive and continuous on [3 ∞). For  ≥ 0,  clearly decreases on [3 ∞); and for   0, it can be verified that  is ultimately decreasing. Thus, we can apply the Integral Test.

 =

 ∞ 3



 ln  [ln(ln )]^ = lim

→∞

  3

[ln(ln )]^−

 ln   = lim

→∞

[ln(ln )]^−+1

− + 1

^

3

[for  6= 1]

= lim

→∞

[ln(ln )]^−+1

− + 1 − [ln(ln 3)]^−+1

− + 1

 ,

which exists whenever − + 1  0 ⇔   1. If  = 1, then  = lim_

→∞

ln(ln(ln ))

3= ∞. Therefore,

∞

=3

1

 ln  [ln(ln )]^converges for   1.

31. Clearly the series cannot converge if  ≥ −¹2, because then lim

→∞(1 + ²)^ 6= 0. So assume   −¹2. Then

 () = (1 + ²)^is continuous, positive, and eventually decreasing on [1 ∞), and we can use the Integral Test.

[continued]

30. Explain why the Integral Test cant be used to determine whether the series is convergent.

∞

X

n=1

cos²n 1 + n² Solution:

⁰() = ⁴+ 1 − 4⁴

(⁴+ 1)² = 1 − 3⁴

 _∞

1



⁴+ 1 = lim

→∞

  1

1 2(2)

1 + (²)²  = lim

→∞

1

2tan⁻¹(²)

 1

= 1 2 lim

→∞[tan⁻¹(²) − tan⁻¹1] = 1 2

  2 −

4

=  8 so the series ^∞

=1



⁴+ 1 converges.

=1

cos 

√ .

=1

cos² 1 + ².

=2

1

 () = 1

 _∞

2

1

(ln )^  = lim

→∞

(ln )¹^−

1 − 

^

2

[for  6= 1] = lim

→∞

(ln )¹^−

1 −  −(ln 2)¹^−

1 − 



30. () = 1

 =

 _∞

3



 ln  [ln(ln )]^ = lim

→∞

  3

[ln(ln )]^−

 ln   = lim

→∞

[ln(ln )]^−+1

− + 1

^

3

[for  6= 1]

= lim

→∞

[ln(ln )]^−+1

− + 1 − [ln(ln 3)]^−+1

− + 1

 ,

which exists whenever − + 1  0 ⇔   1. If  = 1, then  = lim

→∞

3= ∞. Therefore,

∞

=3

1

→∞(1 + ²)^ 6= 0. So assume   −¹2. Then

[continued]

32. Find the values of p for which the series is convergent.

∞

X

n=3

1 n ln n[ln(ln n)]^p Solution:

⁰() = ⁴+ 1 − 4⁴

(⁴+ 1)² = 1 − 3⁴

 _∞

1



⁴+ 1 = lim

→∞

  1

1 2(2)

1 + (²)²  = lim

→∞

1

2tan⁻¹(²)

 1

= 1 2 lim

→∞[tan⁻¹(²) − tan⁻¹1] = 1 2

  2 −

4



=  8 so the series ^∞

=1



⁴+ 1converges.

=1

cos 

√ .

=1

cos² 1 + ².

=2

1

 () = 1

 _∞

2

1

(ln )^ = lim

→∞

(ln )¹^−

1 − 

^

2

[for  6= 1] = lim_

→∞

(ln )¹^−

1 −  −(ln 2)¹^−

1 − 



30. () = 1

 =

 ∞ 3



 ln  [ln(ln )]^ = lim

→∞

  3

[ln(ln )]^−

 ln   = lim

→∞

[ln(ln )]^−+1

− + 1

^

3

[for  6= 1]

= lim

→∞

[ln(ln )]^−+1

− + 1 − [ln(ln 3)]^−+1

− + 1

 ,

which exists whenever − + 1  0 ⇔   1. If  = 1, then  = lim_

→∞

3= ∞. Therefore,

∞

=3

1

→∞(1 + ²)^6= 0. So assume   −¹2. Then

[continued]

46. Use the following steps to show that the sequence tn= 1 +1

2 +1

3+ · · · + 1 n− ln n

has a limit. (The value of the limit is denoted by γ and is called Eulers constant.)

(a) Draw a picture like Figure 6 with f (x) = 1/x and interpret tn as an area [or use (5)] to show that tn > 0 for all n.

(b) Interpret

t_n− tn+1= [ln(n + 1) + ln n] − 1 n + 1 1

(2)

as a difference of areas to show that tn− tn+1> 0. Therefore {tn} is a decreasing sequence.

(c) Use the Monotonic Sequence Theorem to show that {tn} is convergent.

Solution:

136 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 41. ^∞

=1

^−1001 = ^∞

=1

1

^1001 is a convergent -series with  = 1001  1. Using (2), we get

≤

 _∞



^−1001 = lim

→∞

^−0001

−0001

^



= −1000 lim__→∞

 1

^0001





= −1000



− 1

^0001



= 1000

^0001. We want

 0000 000 005 ⇔ 1000

^0001  5 × 10⁻⁹ ⇔ ^0001  1000 5 × 10⁻⁹ ⇔

 

2 × 10¹¹1000

= 2¹⁰⁰⁰× 10^11,000≈ 107 × 10³⁰¹× 10^11,000= 107 × 10^11,301.

42. (a) () =

ln 



2

is continuous and positive for   1, and since ⁰() =2 ln  (1 − ln )

³  0for   , we can apply the Integral Test. Using a CAS, we get

 _∞

1

ln 



2

 = 2, so the series ^∞

=1

ln 



2

also converges.

(b) Since the Integral Test applies, the error in  ≈ ^is ≤

 _∞



ln 



2

 = (ln )²+ 2 ln  + 2

 .

(c) By graphing the functions 1= (ln )²+ 2 ln  + 2

 and 2= 005, we see that 1 2for  ≥ 1373.

(d) Using the CAS to sum the first 1373 terms, we get 1373≈ 194.

43. (a) From the figure, 2+ 3+ · · · + ^≤

1  () , so with

 () = 1

,1 2+1

3+1

4+ · · · + 1

 ≤

  1

1

 = ln .

Thus, = 1 +1 2+ 1

3+1

4+ · · · + 1

 ≤ 1 + ln .

(b) By part (a), ₁₀6 ≤ 1 + ln 10⁶≈ 1482  15 and

₁₀9≤ 1 + ln 10⁹≈ 2172  22.

44. (a) The sum of the areas of the  rectangles in the graph to the right is

1 + 1 2+1

3+ · · · + 1

. Now +1 1



 is less than this sum because the rectangles extend above the curve  = 1, so

 +1 1

1

 = ln( + 1)  1 + 1 2+1

3+ · · · + 1

, and since ln   ln( + 1), 0  1 + 1

2+ 1

3+ · · · + 1

 − ln  = ^. (b) The area under () = 1 between  =  and  =  + 1 is

 +1





 = ln( + 1) − ln , and this is clearly greater than the area of the inscribed rectangle in the figure to the right



which is 1

 + 1

 , so

− ^+1= [ln( + 1) − ln ] − 1

 + 1  0, and so  +1, so {^} is a decreasing sequence.

SECTION 11.4 THE COMPARISON TESTS ¤ 137

(c) We have shown that {^} is decreasing and that ^ 0for all . Thus, 0  ≤ ¹= 1, so {^} is a bounded monotonic sequence, and hence converges by the Monotonic Sequence Theorem.

45. ^{ln }=

^{ln }ln 

=

^{ln }ln 

= ^{ln }= 1

^{− ln }. This is a -series, which converges for all  such that − ln   1 ⇔ ln   −1 ⇔   ⁻¹ ⇔   1 [with   0].

46. For the series ^∞

=1



− 1

 + 1

 ,

=



=1



 − 1

 + 1



=

 1−1

2

 +

 2−1

3

 +

 3−1

4



+ · · · +



− 1

 + 1



= 

1+ − 1

2 + − 1

3 + − 1

4 + · · · + − 1

 − 1

 + 1 =  + ( − 1)

1 2+ 1

3+1

4+ · · · + 1





− 1

 + 1 Thus, ^∞

=1



− 1

 + 1



= lim

→∞= lim

→∞



 + ( − 1)



=2

1

 − 1

 + 1



. Since a constant multiple of a divergent series is divergent, the last limit exists only if  − 1 = 0, so the original series converges only if  = 1.

11.4 The Comparison Tests

1. (a) We cannot say anything about

. If  for all  and

is convergent, then

could be convergent or divergent. (See the note after Example 2.)

(b) If  for all , then

is convergent. [This is part (i) of the Comparison Test.]

2. (a) If  for all , then

is divergent. [This is part (ii) of the Comparison Test.]

(b) We cannot say anything about

. If  for all  and

is divergent, then

could be convergent or divergent.

3. 1

³+ 8  1

³ for all  ≥ 1, so

∞

=1

1

³+ 8converges by comparison with

∞

=1

1

³, which converges because it is a -series with  = 3  1.

4. 1

√ − 1  1

√ for all  ≥ 2, so

∞

=2

√ 1

 − 1 diverges by comparison with

∞

=2

√1

, which diverges because it is a -series with  = ¹₂ ≤ 1.

5.  + 1

√

  

√

 = 1

√ for all  ≥ 1, so^∞

=1

 + 1

√

 diverges by comparison with^∞

=1

√1

, which diverges because it is a p-series with  = ¹₂ ≤ 1.

6.  − 1

³+ 1  

³ = 1

² for all  ≥ 1, so

∞

=1

 − 1

³+ 1converges by comparison with

∞

=1

1

², which converges because it is a -series with  = 2  1.

2