TITLE: convergence of rather uncommon series
QUESTION [5 upvotes]: 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 :)

REPLY [2 votes]: 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".