# Ugly sum

bogdan
_____n__________k+a
lim sum k*(p^k)*C = ?
___k=-a_________k+2*a
n->infinity

0<p<1
a>0

Last edited:

arcnets
Sorry bogdan I can't read it. Maybe better to write it down, scan it and post the image...

bogdan
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 ?

Staff Emeritus
Gold Member
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

bogdan
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...