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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
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Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.

1 # DBsubject('Calculus')
2 # DBchapter('Infinite Series and Sequences')
3 # DBsection('Absolute Convergence and the Root and Ratio Tests')
4 # KEYWORDS('calculus', 'series', 'sequences', 'alternating series', 'convergence', 'absolute convergence')
5 # TitleText1('Calculus: Early Transcendentals')
6 # EditionText1('2')
7 # AuthorText1('Rogawski')
8 # Section1('10.4')
9 # Problem1('19')
10 # Author('LA Danielson')
11 # Institution('The College of Idaho')
12 DOCUMENT();
13
14
15
17
21
22 #Book Values
23 #$series = \sum 1/3^n+5^n #19c, also add in #20d,21c,22d,23c 24 25$a = random(3,7,1);
26 $b =$a+2;
27 $am1 =$a-1;
28
29
30 $an1 = "\frac{1}{$a^n+$b^n}"; 31$an2 = "\frac{n^{$am1}}{n^{$a}-n}";
32 $an3 = "\frac{(-1)^n}{\sqrt{n^2+$a}}";
33 $an4 = "\frac{1}{\sqrt{n^2+$a}}";
34
35 $hn = "\frac{1}{n}"; 36 37 #19c 38$sol1 = "For $$n\ge 1$$,
39
40 $an1 \le \frac{1}{a^n}=\left(\frac{1}{a} \right)^n.$
41
42 The series $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{a} \right)^n$$ is a convergent geometric series, so the Comparison Test implies that the series $$\sum\limits_{n=1}^{\infty} an1$$ converges.";
43
44 #20d
45 $sol2 = "Apply the Limit Comparison Test and compare with the divergent harmonic series: 46 47 $L=\lim_{n\to\infty}\frac{an2}{hn} = 48 \lim_{n\to\infty} \frac{n^{a}}{n^{a}-n}=1 .$ 49 Because $$L>0$$, we conclude that the series $$\sum\limits_{n=1}^{\infty} an2$$ diverges."; 50 51 #21c 52$sol3 = "This is an alternating series with $$a_n = an3$$.  Because $$a_n$$ is a decreasing sequence that converges to zero, the series $$\sum\limits_{n=1}^{\infty} an3$$ converges by the Leibniz Test.";
53
54 #22d
55 $sol4 = "Apply the Limit Comparison Test and compare with the divergent harmonic series: 56 57 $L=\lim_{n\to\infty}\frac{an4}{hn} = 58 \lim_{n\to\infty} \frac{n}{\sqrt{n^2+a}}=1 .$ 59 Because $$L>0$$, we conclude that the series $$\sum\limits_{n=1}^{\infty} an4$$ diverges."; 60 61 ($series, $trueanswer,$solution) = @{list_random(
62   [ "$$\sum\limits_{n=1}^{\infty} an1$$", 'converges',$sol1], 63 [ "$$\sum\limits_{n=1}^{\infty} an2$$", 'diverges',$sol2],
64         [ "$$\sum\limits_{n=1}^{\infty} an3$$", 'converges',$sol3], 65 [ "$$\sum\limits_{n=1}^{\infty} an4$$", 'diverges',$sol4],
66
67 )};
68
69
70
71 #make a multiple choice question
72 $question = new_multiple_choice(); 73$question->qa(' $series ',$trueanswer);
74 $question->makeLast( 'converges', 'diverges'); 75 76 77 Context()->texStrings; 78 79 BEGIN_TEXT 80 \{ beginproblem() \} 81 \{ textbook_ref_exact("Rogawski ET 2e", "10.4", "19") \} 82$PAR
83 Determine convergence or divergence by any method.
84 $PAR 85 \{$question->print_q() \}
86 \{ $question->print_a() \} 87 END_TEXT 88 89 Context()->normalStrings; 90 91 #Answer Check Time! 92 ANS(radio_cmp($question->correct_ans));
93
94 Context()->texStrings;
95 SOLUTION(EV3(<<'END_SOLUTION'));
96 $PAR 97$SOL
98
99 \$solution
100
101 END_SOLUTION
102
103 ENDDOCUMENT()