1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ugly sum

  1. Apr 14, 2003 #1
    lim sum k*(p^k)*C = ?

    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...


    Got it ?
  5. Apr 15, 2003 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    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.

  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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook