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

Question

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 04 '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 04 '13 at 17:26
@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 04 '13 at 17:32
@mjb I believe there is no elementary function with this property — Amr, Jan 04 '13 at 17:38
I edited the original post to state the example more clearly. — mjb, Jan 04 '13 at 17:38
@Amr I could not find one as well, but this is an exercise from a german book. — mjb, Jan 04 '13 at 17:40
@mjb I can find such a function $a_n$ if you were asking about $\frac{x^{2^n}}{1+x^{2^n}}$ — Amr, Jan 04 '13 at 17:43
@Amr Yes, but the exercise states, that it should be done that way. Anyhow: I would not have any other idea to find the limit of this sum. — mjb, Jan 04 '13 at 17:43
@Amr ok. maybe it's a misprint in the book. how would you do that? — mjb, Jan 04 '13 at 17:44
@Mjb Are you sure that you have $-$ in the denominator and not $+$. (check my last comment). — Amr, Jan 04 '13 at 17:44
@Amr In the book it is that way - but as I said: it might be a misprint. Can you give me your idea for the modified (+) term? — mjb, Jan 04 '13 at 17:47
@mjb No. I know how to do it for $\frac{2^nx^{2^n}}{x^{2^n}+1}$. Let me think abt the original problem again — Amr, Jan 04 '13 at 18:01
$x^{2^n}/(1-x^{2^n})$ is not a series. Please edit the title and the body so it asks what you really want to ask. Please don't rely on people wading through all the comments. — Gerry Myerson, Jan 06 '13 at 14:49

Did · Accepted Answer · 2017-05-21T10:52:07.500

2

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

edited May 21 '17 at 10:52

answered Jan 06 '13 at 14:56

Did

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

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