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Induction with sum on left and right side.

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Does
    [itex]\sum_{k=1}^{n}(\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{k=1}^{n}\frac{1}{k + n}[/itex]
    hold for [itex]1 \leq n[/itex]


    2. Relevant equations



    3. The attempt at a solution
    It holds for n = 1. I assume that it should be done with induction but I can't find a way actually compare the two sums to each other. I then had an idea about maybe putting each sum on a common denominator and prove that the both denominators and the both numerators are equal but that's not true so I don't really know where to start.
    The left terms can be rewritten as [itex]\frac{1}{2k(2k-1)}[/itex] as well but that havent really helped me either.
     
  2. jcsd
  3. Mar 18, 2012 #2
    You can prove this using induction. Using the usual induction techniques, define [itex] f(n) = \sum\limits_{k=1}^{k=n}\left(\frac{1}{2k-1} - \frac{1}{2k}\right) [/itex]. You want to show [itex]f(n) = \sum\limits_{k=1}^{k=n} \frac{1}{k+n} [/itex]. You've already shown that this is true for [itex]n=1[/itex]. Assuming the statement is true for [itex]n[/itex]
    [tex]
    f(n+1)\\
    = \sum\limits_{k=1}^{k=n+1}\left(\frac{1}{2k-1} - \frac{1}{2k}\right) = f(n) + \frac{1}{2n+1} - \frac{1}{2n+2} \\
    = \sum\limits_{k=1}^{k=n} \frac{1}{k+n}+ \frac{1}{2n+1} - \frac{1}{2n+2}
    [/tex]
    Now changing the variable in the sum [itex] k \to k' = k-1 [/itex]. The sum becomes
    [tex]
    f(n+1) \\
    = \sum\limits_{k'=0}^{k'=n-1} \frac{1}{k'+n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2} \\
    = \sum\limits_{k'=1}^{k'=n-1} \frac{1}{k'+n+1} + \frac{1}{n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2}\\
    = \sum\limits_{k'=1}^{k'=n-1} \frac{1}{k'+n+1}+ \frac{1}{2n+1} + \frac{1}{2n+2}\\
    =\sum\limits_{k'=1}^{k'=n+1} \frac{1}{k'+n+1}
    [/tex]

    Thus, the statement is true!
     
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