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Telescoping sum

  1. Dec 12, 2006 #1
    1/(16n^2-8n-3)

    I want to find the sum of this series from n=1 to infinity

    By partial fraction I find

    1/(16n^2-8n-3)=1/((4n+3)(4n-1))

    which can be written

    (1/4)(1/(4n-1)-1/(4n+3))

    The sum of this is 1/12, but the right answer is 1/4. What is wrong?
     
    Last edited: Dec 12, 2006
  2. jcsd
  3. Dec 12, 2006 #2

    cristo

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    Well, to start with, your factoring is wrong:

    [tex] 16n^2-8n-3 = (4n+1)(4n-3) [/tex]
     
  4. Dec 12, 2006 #3
    My mistake
     
    Last edited: Dec 12, 2006
  5. Dec 12, 2006 #4

    cristo

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    The term on the RHS is not the expression you give in your question!
     
  6. Dec 12, 2006 #5
    I realized that.

    The simplest mistakes are often the most difficult to find.
     
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