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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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(b) Integrate the result of part (a) from 0 to 1 to get an expression for

1 2 1 2 1 1 3 2 1 4 1 ∙ ∙ ∙ 1 1

2n 2 1 2 1

2n

as an integral.

(c) Deduce from part (b) that

1

Z

1 ? 2 1 1

3 ? 4 1 1

5 ? 6 1 ∙ ∙ ∙ 1 1

s2n 2 1ds2nd 2 y 1

0

dx

1 1 x

Z , y 1

0

x 2n dx

(d) Use part (c) to show that the sum of the given series is ln 2.

21. Find all the solutions of the equation

1 1 x 2! 1 x 2

4! 1 x 3

6! 1 x 4

8! 1 ∙ ∙ ∙ − 0

[Hint: Consider the cases x > 0 and x , 0 separately.]

1

1

1

1

22. Right-angled triangles are constructed as in the figure. Each triangle has height 1 and its

base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes

indefinitely many turns around P by showing that g n is a divergent series.

¨£

¨

¨¡

P

FIGURE for problem 22

1

1

23. Consider the series whose terms are the reciprocals of the positive integers that can be written

in base 10 notation without using the digit 0. Show that this series is convergent and the

sum is less than 90.

24. (a) Show that the Maclaurin series of the function

f sxd −

x

1 2 x 2 x is ò f nx n

2

n−1

where f n is the nth Fibonacci number, that is, f 1 − 1, f 2 − 1, and f n − f n21 1 f n22

for n > 3. [Hint: Write xys1 2 x 2 x 2 d − c 0 1 c 1x 1 c 2x 2 1 . . . and multiply both

sides of this equation by 1 2 x 2 x 2 .]

(b) By writing f sxd as a sum of partial fractions and thereby obtaining the Maclaurin series

in a different way, find an explicit formula for the nth Fibonacci number.

25. Let u − 1 1 x 3

v − x 1 x 4

3! 1 x 6

6! 1 x 9

w − x 2

2! 1 x 5

5! 1 x 8

Show that u 3 1 v 3 1 w 3 2 3uvw − 1.

9! 1 ∙ ∙ ∙

4! 1 x 7

7! 1 x 10

10! 1 ∙ ∙ ∙

8! 1 ∙ ∙ ∙

26. Prove that if n . 1, the nth partial sum of the harmonic series is not an integer.

Hint: Let 2 k be the largest power of 2 that is less than or equal to n and let M be the product

of all odd integers that are less than or equal to n. Suppose that s n − m, an integer. Then

M2 k s n − M2 k m. The right side of this equation is even. Prove that the left side is odd by

showing that each of its terms is an even integer, except for the last one.

790

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