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Ugly sum

  1. Apr 14, 2003 #1
    _____n__________k+a
    lim sum k*(p^k)*C = ?
    ___k=-a_________k+2*a
    n->infinity

    0<p<1
    a>0
     
    Last edited: Apr 14, 2003
  2. jcsd
  3. Apr 14, 2003 #2
    Sorry bogdan I can't read it. Maybe better to write it down, scan it and post the image...
     
  4. Apr 15, 2003 #3
    it's sum from -a to infinity of

    k*(p^k)*(k+2*a)!/[(k+a)!*a!], where k goes from -a to infinity...

    a>0;a->integer;0<p<1;

    Got it ?
     
  5. Apr 15, 2003 #4

    Hurkyl

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    Ok!

    First step is to get rid of that ungainly k coefficient. Use the manipulations:

    k pk = p (k pk-1)
    = p (d/dp) (pk)

    So letting S be the sum of interest:

    S(p) = p (d/dp) &sum pk (k+2a)! / (a! (k+a)!)

    The next step is to shift the sum over to 0..&infin by letting k+a = i:

    S(p) = p (d/dp) &sum pi-a (i+a)! / (a! i!)
    (i = 0 .. &infin)

    Factor out the p-a:

    S(p) = p (d/dp) (p-a &sum pi (i + a)! / (a! i!) )

    What's left is, I think, a fairly standard summation, but I'll be darned if I can remember its value, so I'll have to compute it again:

    T(p) = &sum pi (i+a)! / (a! i!)

    Use the identity:

    (d/dp)a pi+a = (i+a)!/i! pa

    T(p) = (d/dp)a &sum pi+a / i!
    = (d/dp)a (pa &sum pi/i!)
    = (d/dp)a (pa ep)

    So if you know exactly what a is, you can compute T(p) from here, and plug its value into S(p).


    However, I'm 99% sure you can do much better, and that the sum T(p) has a nice closed form solution you can plug into S(p)... I just can't remember it.

    Hurkyl
     
  6. Apr 15, 2003 #5
    Hurkyl...you're a genius...
    (if the solution is correct...because I don't fully understand it...)
    Anyway, thanks...my combinatorics skills are so pathetic...
    Unfortunately...even though I have that result...it's quite complicated to finish it...and I'm not too happy because that sum is strictly related to a...oh...no...that's a good thing...:smile:
     
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