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Where does $\displaystyle\sum_{n=1}^{\infty} \frac{x^{2^n}}{1-x^{2^n}}$ converges to (for some $x \in \mathbb{R}$ with x<1)?

Since we failed with the Ansatz here, we were able to proof that the series diverges for $x>1$ and converges for $x<1$ by the ratio test. But we have no clue to which value it converges. Any ideas?

Did
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mjb
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    Note that $f(x) = f(x^2)+\frac{x^2}{1-x^2}$ – Hagen von Eitzen Jan 06 '13 at 13:56
  • @Raymond Manzoni: Thank you, but this actually is $x^{(2^n)}$ and not $(x^2)^n$, as we would need in the Lambert series – mjb Jan 06 '13 at 14:16
  • @RaymondManzoni This is unrelated. – Did Jan 06 '13 at 14:16
  • @Hagen von Eitzen: Thank you, but I can't verify that. How would that help? What exactly is x? – mjb Jan 06 '13 at 14:19
  • @mjb To verify, simply plug in $x^2$ for $x$ and you will note that the $n$th summand turns into the original $(n+1)$st summand. Therefore, adding the first summand back "repairs" this. Of course $x$ can be any number with $|x|<1$. You don't get an exact expression for $f(x)$, however. – Hagen von Eitzen Jan 06 '13 at 15:09
  • Oups @did is right (I considered $2^n$ as $2n$ sorry...) – Raymond Manzoni Jan 06 '13 at 15:11

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For every $|x|<1$, the sum of the series is $$\sum_{n=1}^{\infty} \frac{x^{2^n}}{1-x^{2^n}}=\sum\limits_{n=1}^{+\infty}\nu_2(n)x^n$$ where, for every nonzero integer $n$, $\nu_2(n)=\max\{k\geqslant0\mid n\in2^k\mathbb Z\}$ is the 2-adic order of $n$.

To show this, expand each denominator as a geometric series with argument $x^{2^n}$. This yields the series as $$ \sum_{n=1}^{+\infty}x^{2^n}\sum_{k=0}^{+\infty}x^{k2^n}=\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty}x^{k2^n}. $$ Now, fix some $i\geqslant1$ and assume that $i=2^\nu j$ with $j$ odd. The term $x^i$ appears in the double sum on the RHS once for each factorization of $i$ as a product $(2^{\nu-1}j)\cdot2$, $(2^{\nu-2}j)\cdot2^2$, ..., and $j\cdot2^{\nu}$, that is, $\nu$ times.

This generating function has no simple form but the associated Dirichlet generating function is $$ \sum_{n=1}^{+\infty}\frac{\nu_2(n)}{n^s}=\frac{\zeta(s)}{2^s-1}. $$

Did
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