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Infinite sequence w/ factorials

  1. Jan 6, 2009 #1
    Hi, I was just looking at this problem with sequences and I was having a question about factorials.

    I understand that the factorials need to get smaller.

    I was just wondering what is the (3)(2)(1) in the problem symbolizing?
    Just that it will eventually reach the end? and what numbers are 3 2 1?
    is it 3n? or -3?

    just getting confused here.

    Its only half the problem, but I understand what needs to be done for the sequence, just the factorial part was the problem.

    heres the pic:

    I appreciate the help.
  2. jcsd
  3. Jan 6, 2009 #2
    Those are the last terms in a factorial... for example let n = 3 then you have have

    (2*3 - 1)* (2*3 - 2) * 3 * 2 *1 = 5*4*3*2*1 = 5! = 120

    Are you familiar with what factorial is?

    Now pick n = 10, 100, 2482842724742824, etc.

    Notice anything?

    Also this isn't really a sequence it's just a limit as n approaches infinity of (2n-1)! it seems.
  4. Jan 6, 2009 #3
    I guess i'm getting confused by the (3)(2)(1)

    is that always going to be there for any n we use? like 3 or 10?

    not sure how it will ever reach 3 2 1 if 2*anything -1 is never 3 2 or 1.

    confused on this part.
  5. Jan 6, 2009 #4
    Yes that's why I asked if you knew the definition of factorial i.e. n! = n*(n-1)*...*1. This definition of factorial works only for nonnegative integers where 0! = 1 by definition. There are other definitions of factorial which allow, for example, factorial of a rational number i.e. (1/2)! but that requires the gamma function I believe. Check out wikipedia for more information.
  6. Jan 7, 2009 #5
    Even if n happens to be, say, 1, in which case 2n - 1 = 1, it's conventional to write (2n - 1)(2n - 2)(2n - 3)...(3)(2)(1) for the (more formal) [itex]\prod_{i=1}^{2n-1} i[/itex] (and it's written (2n - 1)! with the factorial notation). If n happens to be too small to fit the factors such as 3 or 2, then they're just omitted.
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