- #1

- 191

- 0

_____n__________k+a

lim sum k*(p^k)*C = ?

___k=-a_________k+2*a

n->infinity

0<p<1

a>0

lim sum k*(p^k)*C = ?

___k=-a_________k+2*a

n->infinity

0<p<1

a>0

Last edited:

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- Thread starter bogdan
- Start date

- #1

- 191

- 0

_____n__________k+a

lim sum k*(p^k)*C = ?

___k=-a_________k+2*a

n->infinity

0<p<1

a>0

lim sum k*(p^k)*C = ?

___k=-a_________k+2*a

n->infinity

0<p<1

a>0

Last edited:

- #2

- 508

- 0

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

- #3

- 191

- 0

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

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,950

- 19

First step is to get rid of that ungainly k coefficient. Use the manipulations:

k p

= p (d/dp) (p

So letting S be the sum of interest:

S(p) = p (d/dp) &sum p

The next step is to shift the sum over to 0..&infin by letting k+a = i:

S(p) = p (d/dp) &sum p

(i = 0 .. &infin)

Factor out the p

S(p) = p (d/dp) (p

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 p

Use the identity:

(d/dp)

T(p) = (d/dp)

= (d/dp)

= (d/dp)

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

- 191

- 0

(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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