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Tue Nov 8 15:17:41 2011 UTC (18 months, 1 week ago) by aubreyja
File size: 3369 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')
3 ## DBsection('Convergence of Series with positive terms')
4 ## KEYWORDS('calculus', 'infinite series', 'series', 'converge', 'convergence', 'comparison test', 'integral test', 'limit')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('10.3')
9 ## Problem1('12')
10 ## Author('Christopher Sira')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
18 $context = Context(); 19 20$context->variables->add(n=>'Real');
21
22 $a = Real(random(2, 9, 1)); 23$start = Real(random(2, 9, 1));
24
25 $func = "\frac{$a\ln n}{n^2}";
26
27 $f = "$a*ln(x)/x^2";
28 ##$f = Formula("$a*ln(x)/x^2");
29 ##$f->{limits} = [2,4]; 30 31 ##$ans_lim = Real($a/$start*(ln($start)-1)); 32$ans_lim = Real($a/$start*(ln($start)+1)); 33 34$answer = "converges";
35
36 $wrong = "converges"; 37 38 if ($answer eq "converges") {
39     $wrong = "diverges"; 40 } 41 42$mc = new_multiple_choice();
43
44 $mc->qa("the infinite series $$\displaystyle \sum_{n=start}^{\infty} func$$ ", 45$answer);
46 $mc->extra($wrong);
47 $mc->makeLast("diverges"); 48 49 Context()->texStrings; 50 BEGIN_TEXT 51 \{ beginproblem() \} 52 \{ textbook_ref_exact("Rogawski ET 2e", "10.3","12") \} 53$PAR
54 Use the Integral Test to determine whether the infinite series is convergent.
55 $\sum_{n=start}^{\infty} func$
56 Fill in the corresponding integrand and
57 the value of the improper integral.
58 $BR 59 Enter$BBOLD inf $EBOLD for $$\infty$$,$BBOLD -inf $EBOLD for $$-\infty$$, 60 and$BBOLD DNE $EBOLD if the limit does not exist. 61$PAR
62 Compare with
63 $$\int_{start}^{\infty}$$ \{ ans_rule() \} $$dx$$ = \{ ans_rule() \}
64 $PAR 65 By the Integral Test, 66$BR
67 \{ $mc->print_q; \} 68 \{$mc->print_a; \}
69 $PAR 70 END_TEXT 71 Context()->normalStrings; 72 73 ##ANS($f->cmp);
74 ANS(fun_cmp($f,limits=>[2,5])); 75 ANS(num_cmp($ans_lim,strings=>["inf","INF", "-inf","-INF","DNE","dne"]));
76 ANS(str_cmp($mc->correct_ans)); 77 78$j1 = 0;
79
80 Context()->texStrings;
81 SOLUTION(EV3(<<'END_SOLUTION'));
82 $PAR 83$SOL
84 $PAR 85 Let $$f(x) = \frac{a\ln x}{x^2}$$. Because 86 87 $f'(x) = \frac{a(1 - 2 \ln x)}{x^3} ,$ 88$PAR
89 we see that f'(x) < 0 for $$x > \sqrt{e} \approx 1.65$$.  We conclude that f is decreasing on the interval $$x \ge 2$$.  Since f is also positive and continuous on this interval, the Integral Test can be applied.  By Integration by Parts we find
90 $\int \frac{\ln x}{x^2} \, dx = - \frac{\ln x}{x} + \int x^{-2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C ;$
91
92 therefore,
93
94 $\int_{start}^{\infty} \frac{a \ln x}{x^2} \, dx = \lim_{R\to\infty} \int_{start}^{R} \frac{a \ln x}{x^2} \, dx = a \lim_{R\to\infty} \left( \frac{1}{start} + \frac{ \ln start}{start} - \frac{1}{R} - \frac{\ln R}{R} \right) = \frac{a(1 + \ln start)}{start} - a \lim_{R\to\infty} \frac{\ln R}{R}.$
95
96 We compute the resulting limit using L'Hopital's Rule:
97
98 $\lim_{R\to\infty} \frac{\ln R}{R} = \lim_{R\to\infty} \frac{\frac{1}{R}}{1} = 0 .$
99
100 Hence,
101
102 $\int_{start}^{\infty} \frac{a\ln x}{x^2} \, dx = \frac{a(1 + \ln start)}{start} .$
103
104 The integral converges; therefore, the series $$\displaystyle \sum_{n=start}^{\infty} func$$ also converges.  \$PAR
105 END_SOLUTION
106
107 ENDDOCUMENT();
108
109