# convergence of rather uncommon series

While working on my physics bachelor thesis i encountered the following series:

$$\sum_{n=1}^{\infty} \prod_{m=0}^n \frac{x^m}{1-x^{(m+1)}}$$

For $|x|<1$ this sum converges. But i don't know how to get the analytic expression of this result.

I already tried in a rather unsophisticated way to calculate the sum for different values of x between 0 and 1 and fit it to a function. I obtained the best results with the function $f(x)=\frac{a \cdot x}{b + c \cdot x}$ but there is still a significant difference. It looks like that: [deleted since it was the wrong sum]

Does anyone know the result of this sum? Or how to get it?

EDIT: i used the wrong sign in the denominator. it should be $\frac{x^n}{1-x^{(n+1)}}$ like it is now.

EDIT2: I have now enough reputation to post images, thanks :)

EDIT3: Turned out i underestimated my problem. It has actually the form of $$\sum_{n=1}^{\infty} \prod_{m=0}^n \frac{x^m}{1-x^{(m+1)}}$$ I don't even know, if this expressible in elementary functions ... i think i am going to try to approximate it numerically. But if someone finds a solution i'd still be happy to use that :)

• \Pi $\to$ \prod. – Did Jun 26 '13 at 13:58

This a series $\sum\limits_{n=0}^{+\infty}a_nx^n$ with positive integer coefficients $(a_n)_{n\geqslant0}$ and radius of convergence $1$.

An easy upper bound is $a_n\leqslant2^n$ for every $n\geqslant0$. When $n\to\infty$, $a_n\to\infty$. The first $11$ coefficients are $a_0=1$, $2$, $2$, $4$, $4$, $6$, $8$, $10$, $12$, $16$, and $a_{10}=20$.

This is sequence A087135, thus $a_n$ is the number of partitions of $n+1$ where all parts except possibly the two smallest are distinct.

"Unfortunately, there [does not seem to be any known] simple analytic expression for this function".

This is an answer to the first version of the question.

Unfortunately, there is no simple analytic expression for this function. On the other hand, it is related to well know functions.For instance $$F(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^{n+1}} = \frac{L(x)}{x} - \frac{1}{1-x},$$ where $L(x)$ is the Lambert Series $$L(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n}.$$ You can also express it with the $x$-Polygamma function: $$F(x) = \frac{\psi_x(1)}{x\ln x} + \frac{\ln(1-x)}{x\ln x} - \frac{1}{1-x}.$$

• sorry, that i uploaded the wrong version fist. I thought i was far enough to already ask, but i apparently wasn't. I'll think over it longer before i post. Thank you for your effort! – kyra Jun 26 '13 at 14:42