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I want to find the value of this sum.

$$\left(\frac{b}{1-ab}\right)\sum\limits_{n=0}^{\infty}\left(c+ (n+1)I\right)^d\left(\frac{ab}{1-ab}\right)^n $$

My thoughts:
Writing out the first terms

$$S = \frac{b}{1-ab}\left[ (C+I)^d + (C+2I)^d\frac{ab}{1-ab} + (C+3I)^d\left(\frac{ab}{1-ab}\right)^2 ....\right]$$

I would multiply $S$ by $\,\left(\frac{ab}{1-ab}\right)\, $ and some other factor to add $I^d$ to each term in the series. However, I am not sure if this is possible.

$I > 0 $
$0<b<1 $
$0<a<\frac{1}{2} $
$0<d<1 $

share|cite|improve this question
No I is a constant. $I(n+1)$ equals the product of $I$ and $n+1$. – LKL Jan 20 '13 at 2:47
What are all these letters? Of $a, b, c,d, I$, what type of objects are these? Real numbers, positive real numbers, integers, positive integers, or what? – Jonas Meyer Jan 20 '13 at 2:47
Sorry, edited the question. – LKL Jan 20 '13 at 2:49
You have not clearly defined what you mean by solve. If your asking for it to be simplified or changed, then what form do you want it in? – Ethan Jan 20 '13 at 2:50
One does not solve a sum, one sums it or shows it does not converge or something. But one does not solve a sum. – Mariano Suárez-Alvarez Jan 20 '13 at 2:54

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