ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Quiz 3
Solution

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[Version for Division 3]

Given the telescoping series

       4     4     4     4
s  =  --- + --- + --- + --- + ...
      3*1   4*2   5*3   6*4

1. Write down the general term of the series.

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                                 4
                       a   =  -------
                        n     (n+2)*n
   

   With no range of values of n stated, one may assume that the range is   n = 1, 2, 3, ...

   Alternative solutions include

               4
   a   =  -----------  ,   n = 2, 3, 4, ...
    n     (n+1)*(n-1)
   

   and

             4
   a   =  -------  ,   n = 3, 4, 5, ...
    n     n*(n-2)
   

   In both of these cases, the range of values of n must be specified.

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2. Write down and simplify the n^th partial sum   s_n   of the series.

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   Partial fractions (using the cover-up rule):

             4        (4/-2)     (4/2)     -2    2
   a   =  -------  =  ------  +  -----  =  --- + -
    n     (n+2)*n      n+2         n       n+2   n
   
   
            (-2   2)   (-2   2)   (-2   2)   (-2   2)
   ==> s  = ( - + -) + ( - + -) + ( - + -) + ( - + -) + ...
        n   ( 3   1)   ( 4   2)   ( 5   3)   ( 6   4)
   
            (-2    2 )   ( -2     2 )   ( -2    2)
      ... + ( - + ---) + ( --- + ---) + ( --- + -)
            ( n   n-2)   ( n+1   n-1)   ( n+2   n)
   

   All terms cancel (telescope) except +2/1, +2/2, -2/(n+1)   and   -2/(n+2)
   Therefore

   
                                 2     2
                       s  = 3 - --- - ---
                        n       n+1   n+2
   

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3. Hence find the sum   s   of the series.

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                            (      2     2  )    =====
   s  =  lim   s   =  lim   ( 3 - --- - --- ) =  | 3 |
        n-->oo  n    n-->oo (     n+1   n+2 )    =====
   

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[Note:   the version of this quiz for the other divisions is slightly different, but the method of solution is similar.]

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Created 2000 10 26 and modified 2000 10 26 by Dr. G.H. George