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Inductive proof of summation formula

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove by induction the following summation formula:
    [tex]\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}[/tex]
    [tex]n \geq 1[/tex]
    2. Relevant equations
    -
    3. The attempt at a solution
    Inductive step:
    1. [tex]\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}[/tex]

    2. [tex]\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}[/tex]

    To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)

    Thanks in advance!
     
  2. jcsd
  3. Oct 21, 2009 #2

    whs

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    Common denominator!
     
  4. Oct 21, 2009 #3

    Hurkyl

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    You don't need to. All you need to do is to show these two sums of fractions are equal.
     
  5. Oct 21, 2009 #4
    That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks
     
  6. Oct 21, 2009 #5

    Hurkyl

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    As a warmup exercise, prove the following:

    [tex]\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1[/tex]
     
  7. Oct 21, 2009 #6
    see how 1 - 1/n+1 looks first
     
  8. Oct 22, 2009 #7
    I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
     
  9. Oct 22, 2009 #8

    Hurkyl

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    Isn't proving
    [tex]1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}[/tex]​
    exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?
     
  10. Oct 22, 2009 #9
    I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.
     
  11. Oct 22, 2009 #10

    HallsofIvy

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    What do you get when you change
    [tex]1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}[/tex]
    to fractions having the same denominator?

    That's what everyone has been telling you do to. Have you done it?
     
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