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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
File size: 2315 byte(s)
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', 'convergence', 'root test', 'ratio test')
5 # TitleText1('Calculus: Early Transcendentals')
6 # EditionText1('2')
7 # AuthorText1('Rogawski')
8 # Section1('10.5')
9 # Problem1('47')
10 # Author('Emily Price')
11 # Institution('W.H.Freeman')
12 DOCUMENT();
13
14
15
16 #Load Necessary Macros
17
21
22
24
25 #Book Values
26 #exp1 = 3
27 #exp2 = 2
28
29 $exp2 = random(2, 5); 30$exp1 = $exp2 + random(1, 3); 31 32 ($an, $approx,$frac, $term,$answer, $trueanswer,) = @{list_random( 33 [ "\frac{1}{\sqrt[$exp1]{n^{$exp2}-n}}", "\frac{1}{\sqrt[$exp1]{n^{$exp2}}}", "\frac{$exp2}{$exp1}", "\sqrt[$exp1]{\frac{n^{$exp2}}{n^{$exp2} - n}}", "diverges", 'divergent'],
34   [ "\frac{1}{\sqrt{n^{$exp1} - n^{$exp2}}}", "\frac{1}{\sqrt{n^{$exp1}}}", "\frac{$exp1}{2}",  "\sqrt{\frac{n^{$exp1}}{n^{$exp1} - n^{$exp2}}}", "converges", 'convergent'] )}; 35 36 37 38 #Let's try to make a multiple choice question 39$question = new_multiple_choice();
40 $question->qa(' $$\sum\limits_{n=2}^{\infty} an$$ is:',$trueanswer);
41 $question->makeLast( 'convergent', 'divergent'); 42 43 44 Context()->texStrings; 45 46 BEGIN_TEXT 47 \{ beginproblem() \} 48 \{ textbook_ref_exact("Rogawski ET 2e", "10.5", "47") \} 49$PAR
50 Determine convergence or divergence using any method covered so far.
51 $PAR 52 \{$question->print_q() \}
53 \{ $question->print_a() \} 54 END_TEXT 55 56 Context()->normalStrings; 57 58 #Answer Check Time! 59 ANS(radio_cmp($question->correct_ans));
60
61 Context()->texStrings;
62 SOLUTION(EV3(<<'END_SOLUTION'));
63 $PAR 64$SOL
65 This series is similar to a $$p$$-series; because
66 $an \approx approx = \frac{1}{n^{frac}}$
67 for large $$n$$, we will apply the Limit Comparison Test comparing with the $$p$$-series with $$p = frac$$.  Now,
68 $L = \lim_{n \to \infty} \frac{an}{\frac{1}{n^{frac}}} = \lim_{n \to \infty} term = 1.$
69 The $$p$$-series with $$p = frac$$ $answer and $$L>0$$ exists; therefore, the series $$\sum\limits_{n=2}^{\infty} an$$ also$answer.
70
71
72
73 END_SOLUTION
74
75 ENDDOCUMENT()


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