A tricky infinite sum-- solution found numerically, need proof Consider an infinite sum of the following form:
$X Y^{\alpha} + X^2 Y^{\alpha + \alpha^2} +X^3 Y^{\alpha + \alpha^2 + \alpha^3} + ...$
...which can be expressed more succinctly as:
$\sum\limits_{j = 1}^{\infty}X^j \prod\limits_{k = 1}^{j}Y^{\alpha^k}$
...where $0 < X < 1$, $0 < \alpha < 1$, and $Y > 0$. 
Using numerical methods, I have discovered that the solution [Update: not the solution, but a close approximation for some values.] is the following:
$\sum\limits_{j = 1}^{\infty}X^j \prod\limits_{k = 1}^{j}Y^{\alpha^k} = \dfrac{X}{1-X} Y^{\frac{\alpha}{1 - \alpha X}}$
Does anybody know how to prove this analytically? 
Thanks in advance!
Update:
As one commenter showed, the above equality is in fact not a solution but rather an approximation that is very close for some parameter values. So here are some more details:
It seems useful to represent the infinite sum in the following way:
$\sum\limits_{j = 1}^{\infty}X^j \prod\limits_{k = 1}^{j}Y^{\alpha^k} = \dfrac{X}{1-X} \varphi$
...where $\varphi$ is potentially a function of $X$, $Y$, and $\alpha$.
Numerically, I have confirmed the following 2 properties of $\varphi$:


*

*$\lim\limits_{X \to 0} \varphi = Y^\alpha~~$ for $~~Y > 0~~$ and $~~\alpha \in (0,1)$

*$\lim\limits_{X \to 1} \varphi = Y$ for $~~Y > 0~~$ and $~~\alpha = \frac{1}{2}$
I'll accept any answer that moves me substantially closer to a general expression for $\varphi$ or a useful approximation.
 A: I might be missing something, but it seems that 
$$\prod_{k=1}^j Y^{\alpha^k} = Y^{\frac{\alpha (1- \alpha^j)}{1-\alpha}}.$$ So, changing the variable to $$Z=Y^{\frac{\alpha}{1-\alpha}},$$
you get
$$Z\sum_{j=1}^\infty X^j  Z^{-\alpha^j}.$$ This does not appear to simplify further...
A: This is not exactly what you asked for, but its all I have.
Define the following terms:
$$X=\sum_{k=0}^{\infty}x^k \qquad \qquad |x|<1 $$
$$Y=\sum_{r=0}^{\infty} \log\left(y^{\alpha^r} \right) \qquad \qquad 1\le y<e \; \; \text{and} \; \;  |\alpha|<1$$
and now consider these as sets,
$$\top X=\{1,x,x^2,x^3,...,x^k,...\}$$
$$\top Y=\{\log\left(y \right),\log\left(y^{\alpha} \right),\log\left(y^{\alpha^2} \right),\log\left(y^{\alpha^3} \right),...,\log\left(y^{\alpha^r} \right),...\}.$$
Where $\top$ denotes a sum as a sequence, and $\top_{(-s)}$ denotes the sequence with the first $s$ terms taken away.
This is will be an iterative process because in your posed question the power of $Y$ themselves are raised to successively higher powers.
First, compute the following:
$$<\top X,\top Y>=\log\left(y \right)+x\log\left(y^{\alpha} \right)+x^2\log\left(y^{\alpha^2} \right)+x^3\log\left(y^{\alpha^3} \right)+...$$
$$<\top_{(-1)} X,\top Y>=x\log\left(y \right)+x^2\log\left(y^{\alpha} \right)+x^3\log\left(y^{\alpha^2} \right)+x^4\log\left(y^{\alpha^3} \right)+...$$
$$<\top_{(-2)} X,\top Y>=x^2\log\left(y \right)+x^3\log\left(y^{\alpha} \right)+x^4\log\left(y^{\alpha^2} \right)+x^5\log\left(y^{\alpha^3} \right)+...$$
$$<\top_{(-3)} X,\top Y>=x^3\log\left(y \right)+x^4\log\left(y^{\alpha} \right)+x^5\log\left(y^{\alpha^2} \right)+x^6\log\left(y^{\alpha^3} \right)+...$$
$.\\
.\\
.\\$
$$<\top_{(-k)} X,\top Y>=x^{k}\log\left(y \right)+x^{k+1}\log\left(y^{\alpha} \right)+x^{k+2}\log\left(y^{\alpha^2} \right)+x^{k+3}\log\left(y^{\alpha^3} \right)+...$$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$$X\log(y)+x\log(y^\alpha)+x^2(\log(y^\alpha)+\log(y^{\alpha^2}))+x^3(\log(y^\alpha)+\log(y^{\alpha^2})+\log(y^{\alpha^3}))+...$$
$\qquad X\log(y)+x\log(y^\alpha)+x^2\log(y^{\alpha+\alpha^2})+x^3\log(y^{\alpha+\alpha^2+\alpha^3})+...$
The first column contains all the extra terms, luckily it is just $X \log(y)$ so we can subtract this away. Using the magical properties of logarithms we group the rest of terms diagonally, then factor $x^m$ and combine all the sum of logs into multiplication.
Now the question is: how do we calculate the exact value of all these dot products?
If we consider the above table as a grid where $(k,r)$ denote the rows and columns respectively then what we really have is the following summation:
$$\sum_{r=0}^{\infty} \sum_{k=0}^{\infty}x^{r+k}\log(y^{\alpha^r})$$
$$=\sum_{r=0}^{\infty} \sum_{k=0}^{\infty}x^{r+k}\alpha^r\log(y)$$
$$=\log(y)\sum_{r=0}^{\infty} \sum_{k=0}^{\infty}x^{r+k}\alpha^r$$
$$ =\log(y)\sum_{r=0}^{\infty} \sum_{k=0}^{\infty}(x \alpha)^{r}x^{k}$$
$$ =\log(y)\sum_{r=0}^{\infty}(x \alpha)^{r} \sum_{k=0}^{\infty}x^{k}$$
$$ =\log(y)\left(\frac{1}{1-x\alpha}\right)\left(\frac{1}{1-x}\right)$$
Subtracting the extra terms we get,
$$\sum_{k=1}^{\infty}x^k\log(\prod_{s=1}^k(y^{\alpha^s}))=\log(y)\left(\frac{1}{1-x\alpha}\right)\left(\frac{1}{1-x}\right)-log(y)\left(\frac{1}{1-x}\right)$$
$$=\log(y)\left(\frac{1}{1-x}\right)\left(\frac{1}{1-x\alpha}-1\right)$$
$$=\log(y)\left(\frac{1}{1-x}\right)\left(\frac{x\alpha}{1-x\alpha}\right)$$
$$=\log(y^{x\alpha})\left(\frac{1}{1-x}\right)\left(\frac{1}{1-x\alpha}\right)$$
Notes: Since both $X$ and $Y$ are absolutely convergent the work done in the summation is justified. Notice that $|x\alpha|<1$ by the initial restrictions.
A: The below answer is what I have arrived at so far, which will be good enough for what I need it for. I will still accept other answers which can do one of the following:


*

*Provide an explicit general expression.

*Expand on the characterization I develop below, demonstrating additional useful properties.

*Develop a more useful characterization.


My answer:
First, notice that $~~\lim\limits_{\alpha \to 1} \sum\limits_{j = 1}^{\infty} X^j \prod\limits_{k = 1}^{j} Y^{\alpha^k} = \dfrac{XY}{1 - XY}$
This suggests that it might be useful to approach the problem using the usual tricks for geometric sequences, but with a small twist:
$S = XY^{α}+X^2Y^{α+α^2}+X^3Y^{α+α^2+α^3}+...$
$\dfrac{S}{XY^\xi} = Y^{α - \xi}+XY^{α+α^2- \xi}+X^2Y^{α+α^2+α^3- \xi}+...$
$\dfrac{S}{XY^\xi} - Y^{α - \xi} = XY^{α+α^2- \xi}+X^2Y^{α+α^2+α^3- \xi}+...$
$\dfrac{S- XY^{α}}{XY^\xi}  = XY^{α+α^2- \xi}+X^2Y^{α+α^2+α^3- \xi}+...$
$S  = \dfrac{XY^{\alpha}}{1 - X Y^{\xi}}$
...for some $\xi$ which is a function of $\alpha$, $X$, and $Y$, as long as certain technical conditions hold.
The following properties for $\xi$ can be demonstrated:
$0 < \xi < \alpha^2$
$\dfrac{\partial \xi}{\partial \alpha} > 0 $
$\dfrac{\partial \xi}{\partial X} < 0 $
$\dfrac{\partial \xi}{\partial Y} < 0 $
$\lim\limits_{\alpha \to 0} \xi = 0$
$\lim\limits_{\alpha \to 1} \xi = 1$
