View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.3_Convergence_of_Series_with_Positive_Terms/10.3.12.pg

    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();
   14 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   15 loadMacros("PGchoicemacros.pl");
   16 loadMacros("Parser.pl");
   17 loadMacros("freemanMacros.pl");
   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