# Rewrite sequence $\frac{x^{2^n}}{1-x^{2^n}}$ in the form $a_n - a_{n+p}$ where $p \in \mathbb{N}$

For arbitrary $x \in \mathbb{R} \setminus \lbrace -1, 1\rbrace$, how can one rewrite the sequence $\frac{x^{2^n}}{1-x^{2^n}}$ in the form $a_n - a_{n+p}$ where $p \in \mathbb{N}$?

The background is the following: We were able to proof, that $\sum_{n=1}^{\infty} (a_n - a_{n+p}) = \left( \sum_{n=1}^p a_n \right) - pa$, where $a_n \to a$.

So for instance $\sum_{n=1}^\infty \frac{1}{n(n+1)}=1$, since with $a_n := \frac{1}{n}$, I find $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ and $\frac{1}{n} \to 0$ as well as $\sum_{n=1}^1 \frac{1}{n(n+1)}=1$.

So if we could rewrite the original sequence in this form, we would be able to find the limit of the series.

-
So you want to find $a_n$ such that $a_n-a_{n+1}=\frac{x^{2^n}}{1-x^{2^n}}$ – Amr Jan 4 '13 at 17:18
@Amr I think he means $a_n-a_{p+n}=\displaystyle\sum \limits_{n=1}^{\infty}\frac{x^{2^n}}{1-x^{2^n}}$ – user10444 Jan 4 '13 at 17:26
@user10444 I doubt this as the RHS is not a function of $n$ – Amr Jan 4 '13 at 17:28
@Amr: Close. I want to find $a_n$ such that $a_n−a_{n+p}=\frac{x^{2^n}}{1−x^{2^n}}$ with some $p \in \mathbb{N}$. – mjb Jan 4 '13 at 17:32
@Amr My fault I forgot to place $\sum$ before the left side – user10444 Jan 4 '13 at 17:34

Using the 2-adic order function $\nu_2$, one gets $$\frac{x^{2^n}}{1-x^{2^n}}=a_n-a_{n+1},\quad a_n=\sum_{k=1}^{+\infty}\alpha_n(k) x^k,\quad \alpha_n=(\nu_2-n+1)\cdot\mathbf 1_{\nu_2\geqslant n}$$