1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Induction with sum on left and right side.

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    [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]
    = \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}
    Now changing the variable in the sum [itex] k \to k' = k-1 [/itex]. The sum becomes
    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}

    Thus, the statement is true!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook