12pm, Campion 235 Written homework 9

Math 1105: Calculus II (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 9 §8.3(48,60,62,87), §8.4(44,55,57,64) §8.3, Problem 48. Write...
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Math 1105: Calculus II (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 9

§8.3(48,60,62,87), §8.4(44,55,57,64) §8.3, Problem 48. Write the repeated decimal 0.027 = 0.027027 . . . first as a geometric series, then as a fraction. Solution §8.3, 48. Note that .027 = 27 × 10−3 and .000027 = 27 × 10−6 . Continuing this pattern, we have ∞

X 27 27 27 27 0.027 = 0.027027 . . . = 3 + 6 + 9 + . . . = . 10 10 10 103k k=1 In order to find a formula for this sum, we will use the formula for the sum of a geometric series. In order to use the formula for a geometric series, we rewrite the sum:  k ∞ ∞ X X 1 27 = 27 3k 10 103 k=1 k=1  k−1 ∞ X 1 1 27 · 3 = 10 103 k=1  l ∞ X 1 27 = , 1000 1000 l=0 where we’ve substituted l = k − 1 in the last line. (Note that as k goes from 1 to ∞, l goes from 0 to ∞.) Now we have  l ∞ 27 X 27 1 27 27 1 1000 = = = = . 0.027 = 1 1000 1000 1000 − 1 999 37 1 − 1000 l=0 ∞ X

2 , find a formula for the nth term (2k − 1)(2k + 1) k=3 of the sequence of partial sums {Sn }. Then evaluate lim Sn to obtain the value of the series, or state n→∞ that it diverges. §8.3, Problem 60. For the telescoping series

Solution §8.3, 60. We use the method of partial fractions: Suppose that 2 A B = + , (2k − 1)(2k + 1) 2k − 1 2k + 1 so that clearing denominators we have 2 = A(2k + 1) + B(2k − 1) .

Substituting k = 1/2 we find that A = 1, and substituting k = −1/2 we find that B = −1. Thus the partial sum Sn is given by n X k=3

n

 1 1 − 2k − 1 2k + 1 ! ! n n X X 1 1 = − 2k − 1 2k + 1 k=3 k=3     1 1 1 1 1 1 + + ... + − + ... + + = 5 7 2n − 1 7 2n − 1 2n + 1 1 1 = − . 5 2n + 1

X 2 = (2k − 1)(2k + 1) k=3



We have n X k=3

1 1 1 2 = lim Sn = lim − = . n→∞ n→∞ (2k − 1)(2k + 1) 5 2n + 1 5

∞ X √ √ ( k + 1 − k), find a formula for the nth term of §8.3, Problem 62. For the telescoping series k=1

the sequence of partial sums {Sn }. Then evaluate lim Sn to obtain the value of the series, or state that n→∞ it diverges. Solution §8.3, 62. The partial sum Sn is given by n X √ √ ( k + 1 − k) = Sn = k=1

= =

√ √

k=1

! −

n X √ k

!

k=1

 √ √ √ √ √  2 + 3 + ... + n + 1 − 1 + 2 + ... + n

n+1−1 .

Evidently, we have lim Sn = lim n→∞

n X √ k+1

n→∞



n + 1 − 1 = ∞, so the series given diverges.

§8.3, Problem 87. We construct Koch’s snowflake as follows: Let I0 be an equilateral triangle with sides of length 1. The figure I1 is obtained by replacing the middle third of each side of I0 with a new outward pointing equilateral triangle with sides of length 1/3. The process is repeated where In+1 is obtained by replacing the middle third of each side of In with a new outward pointing equilateral triangle with sides of length 1/3n+1 . The limiting figure as n → ∞ is called the snowflake island. (a) Let Ln be the perimeter of In . Show that lim Ln = ∞. n→∞

(b) Let An be the area of In . Find lim An . (Hint: It exists!) n→∞

Solution §8.3, 87 (a). Note that the perimeter of a I0 is 3, since each of its three sides has length 1. To build I1 from I0 we replace each of its three sides with four segments (for a total of 3 · 4 = 12 segments), each of length 1/3 of the previous length 1. This means that the perimeter of I1 is 12 · (1/3) = 4. To obtain I2 from I1 , we replace each of the 12 segments of length 1/3 by four segments of length (1/3) · (1/3) = 1/9. This gives us 3 · 4 · 4 = 48 segments, each of length (1/3)2 .

At the nth step, to obtain the number of segments we multiply 3 · 4n , and each is of length (1/3)n . This implies that the perimeter Ln of In is given by  n  n 1 4 n Ln = 3 · 4 · =3· . 3 3 Since 4/3 > 1, we have  n 4 =∞. lim Ln = lim 3 · n→∞ n→∞ 3 Solution §8.3, 87 (b). Note that the area enclosed by I1 is given by the area enclosed by I0 plus the areas of the three equilateral triangles added to get I1 from I0 . Each of these is an equilateral triangle whose side lengths are exactly 1/3 those of I0 . Scaling a figure in the plane by r scales its area by r2 , so the area of each added triangle is given by (1/3)2 A0 , the total added area is given by 3 · (1/3)2 A0 , and the total area is  2 1 A1 = A0 + 3 · A0 . 3 To obtain the area enclosed by I2 from that obtained by I1 , we add the areas of 3 · 4 equilateral triangles, each with side lengths (1/3)2 . Each of these added triangles has area (1/3)4 A0 , so that total area is given by  2  4 1 1 A2 = A0 + 3 · A0 + 3 · 4 · A0 . 3 3 Similarly, to obtain An , the area enclosed by In , we need to add the areas of each of the equilateral triangles added up to the nth step. The number of sides at the (k − 1)st step is 3 · 4k−1 , and a triangle with side length (1/3)k is added along each of these sides. Each of these triangles has area given by (1/3)2k A0 , so that the area added of the triangles added at the kth step is given by  2k 1 k−1 3·4 · A0 , 3 and we have  2k  k ! n n k X X 1 3 4 3 4 3 · 4k−1 · An = A0 + · 2k A0 = A0 1 + A0 = A0 + . 3 4 3 4 9 k=1 k=1 k=1 n X

The last is almost a geometric series, and we may compute:  k !  k ! n n X 3 4 3 X3 4 lim An = lim A0 1 + = lim A0 1 − + n→∞ n→∞ n→∞ 4 9 4 k=0 4 9 k=1 !  k  k ! n ∞ 1 X3 4 1 X3 4 = lim A0 + = A0 + n→∞ 4 k=0 4 9 4 k=0 4 9       3 1 1 3·9 1 27 8 4 = A0 + = A0 + = A0 + = A0 . 4 4 1− 9 4 36 − 16 4 20 5 In order to finish, we need only compute A0 , the area of the equilateral triangle with side length 1. Drawing an altitude of this equilateral triangle, p one has √ a right triangle with hypotenuse 1 and base p 1/2. Thus the height is given by 1 − (1/2)2 = 3/4 = 3/2. The area A0 can now be computed: √ √ 1 3 3 A0 = · 1 · = . 2 2 4

By our above computation, the area enclosed by Koch’s curve is √ √ 8 3 2 3 · = . 5 4 5

§8.4, Problem 44. Use properties of infinite series to evaluate

∞ X

3e−k .

k=2

Solution §8.4, 44. In order to rearrange the sum to use the formula for a geometric series, we may re-index by letting l = k − 2. Then we have ∞ X

3e

−k

=

∞ X

k=2

3e

−l−2

=

l=0

∞ X

3e−2 e−1

l

.

l=0

The last is in the familiar form for a geometric series, so we have: ∞ X

3e−k =

k=2

§8.4, Problem 55. Determine whether

∞ X k=1

3 3e−2 = . 1 − e−1 e2 − e



k k2 + 1

converges or diverges.

Solution §8.4, 55. Note that we have 1 k = lim q lim √ k→∞ k 2 + 1 k→∞ 1 +

= 1 6= 0 . 1 k2

By the Divergence Test, the given series diverges. §8.4, Problem 57. Determine whether

∞ X k=2

4 converges or diverges. k(log k)2

4 , x log2 x

Solution §8.4, 57. Let f (x) = so that f (k) is the kth term in the series above. By the chain rule we have    1 4 2 2 0 −2 f (x) = −4(x log x) · 1 · log x + x · 2 log x · =− · (log2 x + 2 log x) x x log2 x 4 log x(log x + 2) 4(log x + 2) =− =− . 2 x log x x log x For any value of x > 1, the terms in the fraction are all positive, so f 0 (x) < 0 and f is decreasing. Thus f is a positive, decreasing, continuous function, and the Integral Test applies. In order to determine the convergence or divergence of the given series, we attempt to compute the similar improper integral: Z ∞ Z a 4 dx 4 dx = lim . 2 x log x a→∞ 2 x log2 x 2 The last integral can be computed using the substitution u = log x and du = dx/x: Z Z 4 dx 4 du 4 4 = =− +C =− +C . 2 2 u u log x x log x

Thus we have Z

Z



Since the integral 2



4 dx = lim x log2 x a→∞

Z

a

4 dx x log2 x 2 2  a 4 = lim − a→∞ log x 2 4 4 4 = lim − + = . a→∞ log a log 2 log 2 ∞ X 4 dx 4 converges, we have that the sum converges. 2 x log x k log2 k k=2

§8.4, Problem 64. The prime numbers are those positive integers that are divisible by only 1 and themselves (for example 2, 3, 5, 7, . . .). A celebrated theorem states that the sequence of prime numbers ∞ ∞ X X pk 1 1 {pk } satisfies lim = 1. Show that diverges, which implies that diverges. k→∞ k log k k log k pk k=2 k=1 ∞ X 1 Solution §8.4, 64. In order to check that diverges, we let g(x) = 1/(x log x). By the k log k k=2 chain rule we have   1 + log x 1 0 −2 . f (x) = −(x log x) =− log x + x · x (x log x)2 For x > 2 the terms in the fraction are all positive, so we have f 0 (x) < 0. This implies that f is positive, decreasing, and continuous, and thus the Integral Test applies. In order to compute the improper integral Z ∞ dx x log x 2 we use the substitution u = log x and du = dx/x: Z Z dx du = = log |u| + C = log | log x| + C . x log x u Since x > 2, we have log x > 0, and we have Z dx = log log x + C . x log x Now we may compute: Z ∞ Z a dx dx = lim x log x a→∞ 2 x log x 2 = lim (log log x)|a2 a→∞

= lim log log a − log log 2 = ∞ . a→∞

∞ X dx 1 Since the integral diverges, we have that the sum diverges. As described by the x log x k log k 2 k=2 problem, this implies that the sums of the reciprocals of the prime numbers diverges.

Z