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Closed form for a sequence

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Let Ʃ 1\(n^2-1) from n=2 to k. It says find a closed for it and prove it using sum notation.

    2. Relevant equations



    3. The attempt at a solution
    I can easily prove it by induction but I don't know what a closed form means. I tried looking it up online but there really isn't much info and nothing is stated in my textbook about. All I know is the sequence when expanded looks like

    1/3+1/8+1/15+1/24+...+1/n^2-1. Then not sure how to put it in closed form?
     
  2. jcsd
  3. Jan 29, 2013 #2

    Dick

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    Closed form means find a formula for the sum of series without writing out each term. Try to express 1/(n^2-1) in the form A/(n+1)+B/(n-1). Find A and B. Like partial fractions. Then start writing out terms of that and think about it.
     
    Last edited: Jan 29, 2013
  4. Jan 30, 2013 #3
    This is what I get[tex] \left( \sum_{n=0}^k\frac{1}{-2(n+1)}+\frac{1}{2(n-1)} \right)=\frac{1}{3}+\frac{1}{8}+...+\frac{1}{-2(k+1)}+(\frac{1}{2(k-1)})[/tex]. From here im not sure how to get a closed sum.
     
  5. Jan 30, 2013 #4

    Dick

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    Yes! It's (1/2)*(1/(n-1)-1/(n+1)). Start the sum at n=2 like your original problem posed and bring the (1/2) outside. The series then goes like 1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7... Don't you see a lot of cancellations in there?
     
  6. Jan 30, 2013 #5
    Yes but all I'm left is 1+1\2?
     
  7. Jan 30, 2013 #6

    Dick

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    If you are doing an infinite sum, yes. But if you dealing with the finite sum 2...k you also have some stuff that won't cancel at the k end of the sum.
     
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