Evaluating k-a to k+2a for Limit Sum of k*(p^k)*C

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Discussion Overview

The discussion revolves around evaluating the limit sum of the expression k*(p^k)*C as k approaches -a to k+2a, with specific conditions on the parameters a and p. The focus is on the mathematical manipulation and summation techniques involved in this evaluation, which includes combinatorial aspects.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents the limit sum expression and specifies the conditions 0 < p < 1 and a > 0.
  • Another participant clarifies the summation limits and the factorial terms involved in the expression.
  • A participant proposes a method to simplify the expression by manipulating the k coefficient using differentiation with respect to p.
  • The same participant introduces a transformation of the summation index to shift the limits from -a to infinity, suggesting a standard summation approach.
  • Further, the participant expresses uncertainty about the closed form of the summation T(p) and indicates a need for re-evaluation.
  • Another participant expresses admiration for the proposed method but admits to a lack of understanding and difficulty in completing the evaluation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final evaluation of the limit sum. There are multiple approaches and expressions discussed, but no agreement on a definitive solution or closed form exists.

Contextual Notes

Participants note the complexity of the summation and its dependence on the parameter a, indicating that the resolution may vary based on its value. There are also unresolved steps in the mathematical manipulations presented.

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

0<p<1
a>0
 
Last edited:
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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...

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

Got it ?
 
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
 
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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