View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.2_Summing_an_Infinite_Series/10.2.12.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Limits and Derivatives')
    3 ## DBsection('Definition of the Derivative')
    4 ## KEYWORDS('calculus', 'derivatives', 'slope')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('10.2')
    9 ## Problem1('12')
   10 ## Author('Keith Thompson')
   11 ## Institution('W.H.Freeman')
   12 
   13 DOCUMENT();
   14 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   15 loadMacros("Parser.pl");
   16 loadMacros("freemanMacros.pl");
   17 loadMacros("PGauxiliaryFunctions.pl");
   18 loadMacros("PGgraphmacros.pl");
   19 
   20 #$showPartialCorrectAnswers=1;
   21 
   22 Context()->variables->add(N=>'Real');
   23 
   24 $start=random(4,9,1);
   25 $smm=$start-2;#added for solution, LAD
   26 $sm=$start-1;
   27 $sp=$start+1;
   28 
   29 $S_N = Formula("1/($sm)-1/(N+$sm)");
   30 $ans2=1/($start-1);
   31 
   32 Context()->texStrings;
   33 BEGIN_TEXT
   34 \{ beginproblem() \}
   35 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","12") \}
   36 $PAR
   37 
   38 Write \(S=\sum\limits_{n=$start}^\infty \frac{1}{n(n-1)}\) as a telescoping series and find its sum.
   39 
   40 $PAR \(S_N\) =  \{ans_rule()\}
   41 $PAR \(S\) =  \{ans_rule()\}
   42 END_TEXT
   43 
   44 Context()->normalStrings;
   45 
   46 ANS($S_N->cmp);
   47 ANS(Real($ans2)->cmp);
   48 Context()->texStrings;
   49 SOLUTION(EV3(<<'END_SOLUTION'));
   50 $PAR
   51 $SOL
   52 By partial fraction decomposition
   53 \[\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n},\]
   54 so
   55 \[S=\sum_{n=$start}^\infty \frac{1}{n(n-1)}=\sum_{n=$start}^\infty \left(\frac{1}{n-1}-\frac{1}{n}\right).\]
   56 
   57 The general term in the sequence of partial sums for this series is
   58 \[S_N=\left(\frac{1}{$sm}-\frac{1}{$start}\right) + \left(\frac{1}{$start}-\frac{1}{$sp}\right) + \cdots + \left(\frac{1}{N+$smm}-\frac{1}{N+$sm}\right)=\frac{1}{$sm}-\frac{1}{N+$sm};\]
   59 
   60 thus,
   61 
   62 \[S=\lim_{N\rightarrow \infty} S_N = \lim_{N\rightarrow \infty}\left(\frac{1}{$sm}-\frac{1}{N+$sm}\right)=\frac{1}{$sm}.\]
   63 
   64 END_SOLUTION
   65 
   66 ENDDOCUMENT();