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Factorial identity question

  1. Aug 8, 2012 #1
    [itex]\frac{2 i}{(2 i + 1)!}[/itex] = [itex]\frac{1}{(2 i)!}[/itex] - [itex]\frac{1}{(2 i + 1)!}[/itex]

    Could anybody please show what it is that needs to be done on LHS to get to RHS in this identity.
     
  2. jcsd
  3. Aug 8, 2012 #2
    (2i + 1)! = (2i+1)(2i!)
     
  4. Aug 8, 2012 #3
    [itex]\frac{2i}{(2i+1)!}[/itex]=[itex]\frac{2i}{(2i+1)(2i)! }[/itex]
    =[itex]\frac{2i}{2i(2i)! + (2i)! }[/itex]
    = [itex]\frac{2i}{2i(2i)! + 2i (2i - 1)! }[/itex]
    = [itex]\frac{1}{(2i)! + (2i - 1)! }[/itex]

    and now what?
     
  5. Aug 8, 2012 #4
    Nah, you've gone at that the wrong way.

    Start with the question, and use coolul007's hint to get a common denominator on the right hand side.
     
  6. Aug 8, 2012 #5
    Another way to go about: Add and subtract 1 in the numerator and club 2i and 1.
     
  7. Aug 8, 2012 #6
    thanks acabus, all
     
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