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Method of differences

  1. May 26, 2013 #1
    find [tex] \displaystyle\sum_{r=1}^n (r-1 + \dfrac{1}{r} - \dfrac{1}{r+1}) [/tex]

    using the method of differences

    if I split it up (i.e do sum of r and sum of 1 and sum of (1/r - 1/(r+1)) to n I get the right answer of [itex] \dfrac{n(n^2 + 1)}{2(n+1)} [/itex]

    however if I do the method of differences right away of the expression I get:

    r= 1: 0 + 1 - 1/2
    r= 2: 1 + 1/2 - 1/3
    r= 3: 2 + 1/3 - 1/4
    ...

    r= n-1: n-2 + 1/(n-1) - 1/n
    r = n: n-1 + 1/n - 1/(n+1)

    and adding all of this I get [tex] \displaystyle\sum_{r=0}^{n-1} r - \dfrac{1}{n+1} [/tex]

    = [tex] \dfrac{n(n-1)}{2} - \dfrac{1}{n+1} [/tex]

    which doesn't give me the right answer...

    where have I gone wrong using the second method?
     
  2. jcsd
  3. May 26, 2013 #2

    Ray Vickson

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    Homework Helper

    You dropped a "+1". You should have gotten
    [tex] 1 + \frac{n(n-1)}{2} - \frac{1}{n+1} [/tex]
     
  4. Jun 19, 2013 #3
    This is an telescoping Sequence
     
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