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

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In summary, the conversation involves a calculation of the sum from -a to infinity of k*(p^k)*(k+2*a)!/[(k+a)!*a!], where k goes from -a to infinity. The first step is to get rid of the k coefficient by using manipulations. The next step is to shift the sum over to 0 to infinity by letting k+a = i. Then, the sum is factored to a standard summation, which can be computed by using an identity. However, a closed form solution for the sum is needed to complete the calculation, which may be complicated due to its relation to a.
  • #1
bogdan
191
0
_____n__________k+a
lim sum k*(p^k)*C = ?
___k=-a_________k+2*a
n->infinity

0<p<1
a>0
 
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  • #2
Sorry bogdan I can't read it. Maybe better to write it down, scan it and post the image...
 
  • #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 ?
 
  • #4
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
 
  • #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:
 

1. What is the significance of evaluating k-a to k+2a for Limit Sum of k*(p^k)*C?

The evaluation of k-a to k+2a for Limit Sum of k*(p^k)*C is important in determining the convergence or divergence of the series. It helps in understanding the behavior of the series as the value of k approaches infinity.

2. How is the value of k determined in this evaluation?

The value of k is typically determined by setting it equal to a variable, such as n, and taking the limit as n approaches infinity. This allows us to analyze the behavior of the series as the number of terms increases.

3. What does the variable a represent in this evaluation?

The variable a represents the difference between consecutive terms in the series. It is used to determine the range of values for k that will be evaluated.

4. In what situations would evaluating k-a to k+2a for Limit Sum of k*(p^k)*C be useful?

This evaluation is useful when studying infinite series in mathematics and physics. It can also be used in real-life situations where a quantity increases or decreases over time, such as population growth or radioactive decay.

5. Are there any limitations to this evaluation method?

One limitation of this evaluation method is that it only works for series where the ratio of consecutive terms has a limit. It may also be challenging to determine the value of a in some cases. Additionally, this method may not work for series with complex or non-linear terms.

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