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Thomas Calculus 13th [Solutions]

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Section 10.2 Infinite Series 719<br />

87. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series<br />

that diverges does not change the divergence of the series.<br />

88. Let An<br />

a1 a2<br />

a n and lim An<br />

A . Assume an b n converges to S. Let<br />

n<br />

Sn a1 b1 a2 b2 an bn Sn a1 a2 an b1 b2<br />

bn<br />

b1 b2 bn Sn An lim b1 b2<br />

bn S A b n converges. This contradicts the<br />

n<br />

assumption that b n diverges; therefore, an b n diverges.<br />

89. (a)<br />

(b)<br />

2 5 2 1 r r 3 ; 2 2 3 2 3<br />

1 r 5 5 5 5<br />

13<br />

2<br />

13 3 13 13 3 13 3<br />

2<br />

13 3<br />

3<br />

5 1 r r ;<br />

1 r 10 10 2 2 10 2 10 2 10<br />

2<br />

90.<br />

91.<br />

b 2b 1 1 b b<br />

1 e e 9 1 e e 8 b ln 8<br />

b<br />

1 e 9 9 9<br />

2 3 4 5 2n<br />

2n<br />

1<br />

sn<br />

1 2r r 2r r 2r r 2 r , n 0, 1,<br />

2 4 2n 3 5 2n 1<br />

s 1 2 2 2 2 lim 1 2r 1 2r<br />

n n r r r r r r r s<br />

,<br />

2 2 2<br />

n 1 r 1 r 1 r<br />

2<br />

if r 1 or r 1<br />

92. area 2 2<br />

2<br />

2 2 (1) 2 1 4 2 1 1 4 2<br />

2 1<br />

8 m 1<br />

2<br />

( 0.12)(24)<br />

93. (a) After 24 hours, before the second pill: 300e<br />

16.840 mg; after 48 hours, the amount present<br />

after 24 hours continues to decay and the dose taken at 24 hours has 24 hours to decay, so the amount<br />

( 0.12)(48) ( 0.12)(24)<br />

present is 300e<br />

300e<br />

0.945 16.840 17.785 mg.<br />

2<br />

(b) The long-run quantity of the drug is<br />

( 0.12)(24)<br />

300 ( 0.12)(24)<br />

n<br />

300 e<br />

e<br />

( 0.12)(24)<br />

17.84 mg.<br />

1<br />

1 e<br />

94.<br />

a 1 r n<br />

L s a<br />

ar<br />

n 1 r 1 r 1 r<br />

n<br />

95. (a) The endpoint of any closed interval remaining at any stage of the construction will remain in the Cantor<br />

set, so some points in the set include 0,<br />

1<br />

,<br />

2<br />

,<br />

1<br />

,<br />

2<br />

,<br />

7<br />

,<br />

8<br />

,<br />

1<br />

,<br />

2<br />

,<br />

7<br />

,<br />

8<br />

,1.<br />

(b) The lengths of the intervals removed are:<br />

1<br />

Stage 1:<br />

3<br />

Stage 2: 1 1<br />

1 2<br />

3 3 9<br />

Stage 3: 1 1 2 4<br />

1<br />

3 3 9 27<br />

and so on.<br />

27 27 9 9 27 27 3 3 9 9<br />

Thus the sum of the lengths of the intervals removed is<br />

n 1<br />

1 2 1 1<br />

3 3 3 1 (2 / 3)<br />

n 1<br />

1.<br />

Copyright<br />

2014 Pearson Education, Inc.

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