lim sum k*(p^k)*C = ?
Sorry bogdan I can't read it. Maybe better to write it down, scan it and post the image...
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 ?
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...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...
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