# 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?

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 $\varphi$:

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

2. $\lim\limits_{X \to 1} \varphi = Y$ 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.

• What have you tried? Dont think it is as difficult as you think. Maybe i am mistaken... – Lost1 Dec 28 '13 at 19:11
• Stated in current form, your equality is wrong. With $x = y = \alpha = 1/2$, RHS of your equality is $2^{-2/3} \sim 0.62996052494744$ while the sum of first ten terms in LHS is $\sim 0.63421166281522$, already bigger than RHS. – achille hui Dec 28 '13 at 20:06
• I do not see how X could become part of the exponent of Y. – Claude Leibovici Dec 29 '13 at 6:00
• I checked using $X = Y = \alpha = \frac{1}{2}$ and achille hui is correct. I arrived at the equality using other numbers for which the equality is a very, very close approximation. For $X = Y = \alpha = \frac{1}{2}$ it apears to be a close-ish approximation, but not quite. Rats. – Matt D. Dec 29 '13 at 8:25

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...

• This is a useful representation, thanks. – Matt D. Dec 29 '13 at 8:29

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.

• I'm upvoting this for effort but unfortunately it doesn't seem very useful with logged y's. Thanks. – Matt D. Dec 29 '13 at 14:20

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:

1. Provide an explicit general expression.
2. Expand on the characterization I develop below, demonstrating additional useful properties.
3. Develop a more useful characterization.

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:

$0 < \xi < \alpha^2$

$\dfrac{\partial \xi}{\partial \alpha} > 0$

$\dfrac{\partial \xi}{\partial X} < 0$

$\dfrac{\partial \xi}{\partial Y} < 0$

$\lim\limits_{\alpha \to 0} \xi = 0$

$\lim\limits_{\alpha \to 1} \xi = 1$