If ∑ kr k converges, so will r kr k ∑ , by

48. If ∑ kr k converges, so will r kr k ∑ , by

45. (Proof by contradiction) Assume ∑ ( a k + b k ) Theorem B.

rS = r ∑ kr k = k + 1 ∑ kr = ∑ ( – 1) k r k while

converges. Since

∑ k b converges, so would = 1 = 1 k = 2

kr ∑ k =+ r ∑ kr k so

∑ a k = ∑ ( a k + b k ) ( 1) +− b k , by

– k rS =+ r ∑ kr

– ∑ ( – 1) k r

Theorem B(ii).

∞ 1 =+ r

∑ [ – ( – 1)] k k r =+ r ∑ r = ∑ r

46. (Answers may vary). a n =

and

∑ thus

1 r Since

b n = ∑ ( 1) − both diverge, but

1– r

⎜ n n ⎟ converges to 0.

47. Taking vertical strips, the area is

− nkt

49. a. A = ∑ Ce = ∑ C ⎜ kt ⎟

2 4 8 k ⎝⎠

C Ce = kt 1 2

Taking horizontal strips, the area is

2 4 8 ∑ k 16 . k = 1 2 1 ln 2

a. ∑ k = ∑ ⎜⎟ = 1 = 2 if C = 2 mg, then A = mg.

k = 1 2 k = 1 ⎝⎠ 2 1 −

b. The moment about x = 0 is

∑ ⎛⎞⋅ ⎜⎟ (1) k = ∑ k = ∑ k = 2.

k = 0 ⎝⎠ 2 k = 0 2 k = 1 2

moment 2 x =

area

50. Using partial fractions,

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51. = k + 2 – – f k =

ff kk + 1 f k + 1 f k + 2 ff kk + 1 f k + 2 ff kk + 2 since f k + 2 = f k + 1 + f k . Thus,

The terms of the Fibonacci sequence increase without bound, so

9.3 Concepts Review

4. is continuous, positive, and

2 x 2 bounded above + 1

nonincreasing on [1, ) ∞ .

2. f(k); continuous; positive; nonincreasing

3. convergence or divergence

= 3 π − − tan ⎜ 1 2 ⎟ <∞ 2 ⎝ 2 ⎠

4. p>1

Problem Set 9.3

The series converges.

1. is continuous, positive, and nonincreasing

5. is continuous, positive, and

on [0, ) ∞ .

nonincreasing on [1, ) ∞ .

dx =⎡

0 x + 3 ⎣ ln x +⎤=∞ 3 ⎦ 0 – ln 3 =∞ dx = ∫ ⎡ ⎣ 4 x + 2 1 ⎤ x + 2 ⎦ =∞ –43 =∞ 1

The series diverges.

Thus ∑

diverges, hence

k is continuous, positive, and nonincreasing = 1 k + 2

also diverges.

is continuous, positive, and nonincreasing on [100, ) ∞ .

The series diverges.

( x + 2) 2

3. 2 is continuous, positive, and nonincreasing

2 dx = ⎢ –

) ⎣ . ( x + 2) [ ⎦ 100

The series converges.

The series diverges.

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7. is continuous, positive, and

is continuous, positive, and

x (ln ) x 2

nonincreasing

nonincreasing on [5, ) ∞ .

2 4 x + 2 ⎢ ln 4 x + 2 ⎥ =∞ – ln10 =∞ x (ln ) x

⎣ ln x ⎦ 5 ln 5

The series diverges.

ln 5

The series converges.

8. e x

is continuous, positive, and nonincreasing

[2, ) ∞ . Using integration by parts twice, with

13. lim

= lim

k =≠ 1 0, so the series k = →∞ i k 2 + 5 k →∞ 1 + 5

i = 1, 2 and dv = e – x dx , k 2

xe 2– x dx = [– xe 2– x ] ∞

2 ∫ xe dx 2 2

diverges.

2– x

[– xe ] 2 ⎜ 2 [– xe ] ∞ + ∫ e – 2 x dx ⎞

⎜⎟ ; a geometric series with

= [– xe 2– x –2 xe – x –2 e – x ] ∞ 2

a = 3 ,; r = 3 3 < 1 so the series converges.

The series converges.

15. ∑ ⎜⎟ is a geometric series with r = ;1 <

⎛⎞ k 1 11

k = 1 ⎝⎠ 2 22 (4 3 ) + x 7/6

9. is continuous, positive, and

so the series converges.

nonincreasing on [1, ) ∞ .

In ∑

6 the series diverges. Thus, the sum of the series =+ 0 1/ 6 =⋅ 67 –1/ 6 <∞ diverges.

7 The series converges.

16. is continuous, positive, and nonincreasing on

1000 x 2

10. is continuous, positive, and

[1, ) ∞ . ∫ dx = – =+=<∞ 011 , so

nonincreasing on [2, ) ∞ .

∫ converges.

ln 1 ⎢ 3 + x ⎥ ∑ 2

1 k ∞ ⎛⎞ 1

=∞ 1000 – ln 9 =∞

∑ k = ∑ ⎜⎟ ; a geometric series with

k = 1 2 k = 1 ⎝⎠ 2

The series diverges.

11 r = ; 1, < so the series converges. Thus, the

11. xe –3x is continuous, positive, and

sum of the series converges.

nonincreasing on [1, ) ∞ .

3 <∞ 6e The series converges.

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18. As k →∞ , 0. → Let y = , then

17. sin ⎜ ⎟ ⎞= ⎨ –1 k = 4 j + 3, k

⎩ 0 k is even 1 1 sin lim y k sin = lim sin y = lim =≠ 1 0, so

where j is any nonnegative integer.

k →∞

the series diverges.

Thus lim sin ⎜ ⎟ does not exist, hence

k →∞

xe 2–x is continuous, positive, and

⎞≠ 0 and the series diverges.

lim sin

nonincreasing on [1, ) ∞ .

the series converges.

The series converges to 1.

tan –1 x

2 is continuous, positive, and

is continuous, positive, and

xx

x 3/2

nonincreasing on [1, ) ∞ .

nonincreasing on [5, ) ∞ .

tan –1 x

⎥ ∫ E x ) = 1 2 =⎢ ∑ ≤

, so the series

is continuous, positive, and nonincreasing

is continuous, positive, and

2 dx [tan –1 x ] ∞ 5 nonincreasing on [1, ) ∞ .

1 4x

= – tan –1 5 ≈ 0.1974 ∫

2 26. xx is continuous, positive, and ( + 1)

so the series converges.

nonincreasing on [5, ) ∞ .

23. x is continuous, positive, and nonincreasing on

dx = ∫ ⎜ –

5 0 – ln ∑ x + 1

=⎡ ⎣ ln x – ln x +⎤= 1 ⎦ ⎢ ln

k ≤ ∫ x dx = [– xe ] 5 + ∫ e dx ⎣

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27. E n

∑ 2 ∫ n 2 dx = lim ∫

32. E n = ∑

A →∞ ∫ − n ⎜ + ⎟ dx = ⎝ x x 1 ⎠

⎟ ⎞< 0.0002 ⇒+< 1 e 0.0002 ≈ 1.0002 lim − =

∑ 2 ∫ n 2 dx

1 = ∞ < 1 33. Consider

∫ 2 x p (ln ) x

A →∞ ⎣

p dx ∫

tan n

du which converges for

2 x (ln ) x

ln 2 u p

π − − 1 tan n < 0.0002 ⇒

− tan 1

n >− π 0.0002 p 2 > 1. 2

⇒> n tan π ( 2 − 0.0002 ) ≈ 5000

34. is continuous, positive, and x ln ln(ln ) x x

30. E n =

dx = nonincreasing on [3, ) ∑ ∞.

k 2 ∫ n kn 2 =+ 1 e e x

u = x 2 ∫ 3 x ln ln(ln ) x x

Let u = ln(ln x), du =

dx .

lim

A →∞ 2 ∫ n e u

x ln x

dx = ∫

du = [ln ] u ln(ln 3)

⎜ − ⎟ lim ⎢ A −

3 x ln ln(ln ) x x

ln(ln 3) u

⎝ 2 ⎠ A →∞ 2 ⎣ 2 ⎢ e

=∞ – ln(ln(ln 3)) =∞

2 < 0.0002 ⇒> n ln

1 1 The series diverges.

31. E n = ∑

1 A du

4 dx = 1 lim + k 1 + x A →∞ 2 ∫ 2 1 2

⎣ ⎢ − tan ()

22 ⎢ tan ⎣ ()

The upper rectangles, which extend to n + 1 on

n ⎥ < 0.0002

the right, have area 1 + + +…+ . These

⇒− − 2 tan () n 2 <

⇒ − 1 0.0004 2 tan n > 1.5703963

rectangles are above the curve y = from x = 1

⇒> n tan 1.5703963 ( ) ≈ 50 to x = n + 1. Thus,

n + 1 ∫ 1 dx = x n

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1 1 1 b. < + + +…+ The leftmost rectangle has area

1 ⋅ f (1) = f (1). If each shaded region to the

The lower (shaded) rectangles have area

right of x = 2 is shifted until it is in the

leftmost rectangle, there will be no overlap

+ +…+ . These rectangles lie below the

of the shaded area, since the top of each rectangle is at the bottom of the shaded

1 curve y = from x = 1 to x = n. Thus

region to the left. Thus, the total shaded area

is less than or equal to the area of the

1 1 1 n 1 leftmost rectangle, or B ≤ f (1).

+ +…+ < ∫ dx = n so

ln , n

c. By parts a and b, { B n } is a nondecreasing

1 + + +…+ < + 1 ln . n sequence that is bounded above, so lim B n

n →∞

exists.

36. From Problem 35, B n is the area of the region

d. Let fx () = , then

within the upper rectangles but above the curve

= y 1 . Each time n is incremented by 1, the

∫ () = 1 f x dx ∫ dx = ln( n + 1) 1 and 1 x

added area is a positive amount, thus B n is

lim B n = as defined in Problem 37. γ

increasing.

n →∞

From the inequalities in Problem 35,

41. Every time n is incremented by 1, a positive

< + + +…+ 1 1 01 1 – ln( n +<+ 1) 1 ln – ln( n n + 1)

amount of area is added, thus { A n } is an

increasing sequence. Each curved region has

=+ n

horizontal width 1, and can be moved into the

1 ln

heavily outlined triangle without any overlap.

This can be done by shifting the nth shaded

Since

< 1, ln

< 0, thus B < for all n, 1 region, which goes from (n, f(n)) to

(n + 1, f(n + 1)), as follows:

and B n is bounded by 1.

shift (n + 1, f(n + 1)) to (2, f(2)) and (n, f(n)) to (1, f(2)–[f(n + 1) – f(n)]).

37. { B n } is a nondecreasing sequence that is

The slope of the line forming the bottom of the

bounded above, thus by the Monotonic Sequence

shaded region between x = n and x = n + 1 is

Theorem (Theorem D of Section 9.1), lim B n fn ( + 1) – ( ) fn

exists. The rationality of γ is a famous unsolved since f is increasing.

problem.

By the Mean Value Theorem,

fn ( + 1) – ( ) fn = fc ′ () for some c in (n, n + 1).

38. From Problem 35, ln( n +< 1) ∑ <+ 1 ln , n thus

Since f is concave down, n < c < n + 1 means that

fc ′ () < fb ′ () for all b in [1, n]. Thus, the nth 10,000,000 1 shaded region will not overlap any other shaded

ln(10, 000, 001) 16.1181 ≈ <

region when shifted into the heavily outlined

triangle. Thus, the area of all of the shaded <+ 1 ln(10, 000, 000) 17.1181 ≈ regions is less than or equal to the area of the

39. γ + ln( n +> 1) 20 ⇒ ln( n +> 1)

20 – γ ≈ 19.4228 n n →∞

heavily outlined triangle, so lim A exists.

⇒+> n 1 e 19.4228 ≈ 272, 404,867 ⇒> n 272, 404,866

40. a. Each time n is incremented by 1, a positive

amount of area is added.

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42. ln x is continuous, increasing, and concave down on [1, ) ∞ , so the conditions of Problem 41 are met.

n a. See the figure in the text for Problem 41. The area under the curve from x = 1 to x = n is ∫ ln

area of the nth trapezoid is

, thus A n = ∫ 1 ln x dx – ⎢

Using integration by parts with u = ln x, du = 1 dx , dv = dx, v = x x

∫ n ln

x dx = [ ln ] – x x n 1 ∫ dx = [ ln – ] x x x n

= n ln – – (ln1 –1) n n = n ln – n n + 1

The sum of the areas of the n trapezoids is ln1 ln 2 + ln 2 ln 3 +

= ln(2 3 ⋅ ⋅…⋅ n )–

= ln !– ln n n

Thus, A n = n ln – n n + 1 – ln !– ln ( n n ) = n ln – n n + 1 – ln ! ln n + n = ln n n n – ln e + 1 – ln ! ln n + n

b. By Problem 41, lim A n exists, hence part a says that lim 1 ln ⎢ +⎢

⎥=+⎢ 1 ln lim ⎜⎟

Since the limit exists, lim

n →∞ ⎜⎟ ⎝⎠ e n !

= m . m cannot be 0 since lim ln x =∞ –.

Thus, lim

= lim

= , i.e., the limit exists.

n →∞ n

lim () n e

n n →∞ n n

() n

n →∞ () e n !

⎛⎞ 15 n ⎛ 15 ⎞

c. From part n b, ! ≈ 2 π⎜⎟ n , thus, 15! ≈ 30 π ⎜ ⎟ ≈ 1.3004 10 × 12 ⎝⎠ e ⎝ e ⎠

The exact value is 15! 1,307, 674,368, 000 = .

43. (Refer to fig 2 in the text). Let b k = ∫ () f x dx

; then from fig 2, it is clear that k ≥ b k for k = 1, 2, …… ,, n

Therefore t ∑

a k ≥ ∑ b k = ∫ f x dx

so that

E n = ∑ a k = lim ∑ a k ≥ lim ∫ f x dx =

t →∞ n + 1 ()

∫ n + 1 () .

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9.4 Concepts Review

The series converges.

4. Ratio; Limit Comparison

lim

Problem Set 9.4

n →∞ 1 ++ 5 10 n 10 2 + 3 + 5 1

1. a n =

1 The series diverges.

= ∞ since lim ⎞= 1

⎝ n () ⎠

n →∞ n + 1 99 n →∞ ⎜ ⎟

b n diverges ⇒ ∑ a n diverges.

n = 1 n = 1 The series diverges.

a n + 1 ( n + n 1) n ()

The series converges.