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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')
    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('76')
   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 
   23 $start = 2;
   24 
   25 $func = "\frac{1}{n^a \ln n}";
   26 
   27 @incorrect = ("\( a \le 1 \)",
   28               "\( a \ge 1 \)",
   29               "\(a=0\)",
   30               "\( 0< a < 1\)",
   31               "\(a>0\)",
   32               "\(a \ge 0\)",
   33               "\( a < 1 \)");
   34 
   35 
   36 $mc = new_multiple_choice();
   37 
   38 $mc->qa("For which \(a\) does \( \displaystyle \sum_{n=$start}^{\infty} $func \) converge?",
   39     "\( a > 1 \)");
   40 $mc->extra(@incorrect[NchooseK(7,3)]);
   41 
   42 Context()->texStrings;
   43 BEGIN_TEXT
   44 \{ beginproblem() \}
   45 \{ textbook_ref_exact("Rogawski ET 2e", "10.3","76") \}
   46 $PAR
   47 \{ $mc->print_q; \}
   48 \{ $mc->print_a; \}
   49 $PAR
   50 END_TEXT
   51 Context()->normalStrings;
   52 
   53 ANS(str_cmp($mc->correct_ans));
   54 
   55 $j1 = 0;
   56 
   57 Context()->texStrings;
   58 SOLUTION(EV3(<<'END_SOLUTION'));
   59 $PAR
   60 $SOL
   61 $PAR
   62 First consider the case \(a > 1\).  For \( n \ge 3 \), \( \ln n > 1 \) and
   63 
   64 \[ \frac{1}{n^a \ln n} < \frac{1}{n^a} .\]
   65 
   66 The series \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^a} \) is a \(p\)-series with \(p = a > 1\), so it converges; hence, \( \displaystyle \sum_{n=3}^{\infty} \frac{1}{n^a} \) also converges.  By the Comparison Test we can therefore conclude that the series \( \displaystyle \sum_{n=3}^{\infty} \frac{1}{n^a \ln n} \) converges, which implies the series \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{n^a \ln n} \) also converges.
   67 $PAR
   68 For \( a \le 1 \), \(n^a \le n \) so
   69 
   70 \[ \frac{1}{n^a \ln n} \ge \frac{1}{n \ln n} \]
   71 
   72 for \( n \ge 2 \).  Let \( f(x) = \frac{1}{x \ln x}\).  For \( x \ge 2 \), this function is continuous, positive and decreasing, so the Integral Test applies.
   73 $PAR
   74 Using the substitution \( u = \ln x \), \( du = \frac{1}{x} \, dx \), we find
   75 
   76 \[ \int^{\infty}_{2} \frac{dx}{x \ln x} =  \lim_{R\to\infty}  \int^{R}_{2} \frac{dx}{x \ln x} = \int^{\ln R}_{\ln 2} \frac{du}{u} = \lim_{R\to\infty} (\ln (\ln R) - \ln (\ln 2)) = \infty .\]
   77 The integral diverges; hence, the series \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) also diverges.  By the Comparison Test we can therefore conclude that the series \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{n^a \ln n} \) diverges.
   78 $PAR
   79 To summarize,
   80 \[ \sum_{n=2}^{\infty} $func \text{ converges for } a > 1 \text{ and diverges for } a \le 1 \]
   81 $PAR
   82 END_SOLUTION
   83 
   84 ENDDOCUMENT();
   85 
   86 

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