# Closed form expression for a formal power series

Is there a closed form expression for the following formal power series $$\large\sum_{k\ge 0}\dfrac{z^{2^k}}{1+z^{2^k}}$$ Till now I have tried in vain finding any progress to simplify this expression. Please help.

• what is the limit of $k$??? infinity or not??? – E.H.E Sep 16 '15 at 20:39
• I thought $k\ge 0$ implies $0\le k<\infty$. Is that not standard? – Samrat Mukhopadhyay Sep 16 '15 at 20:41
• @SamratMukhopadhyay It does. Your notation is fine. – Mark Viola Sep 16 '15 at 20:43
• Hmm. Thanks @Dr.MV. – Samrat Mukhopadhyay Sep 16 '15 at 20:43
• The equivalent problem is to evaluate the series $$\sum_{k=0}^\infty \frac{1}{1+x^k}$$for $|x|>1$. WA gives a result that depends on the $x$-digamma function $\psi_x$. So, if you consider that closed-form enough, then you have a way forward. – Mark Viola Sep 16 '15 at 20:52

If we set: $$f(z)=\sum_{k\geq 0}\frac{z^{2^k}}{1+z^{2^k}}\tag{1}$$ we have: $$f(z^2)=\sum_{k\geq 0}\frac{z^{2\cdot 2^k}}{1+z^{2\cdot 2^k}}=\sum_{k\geq 0}\frac{z^{2^{k+1}}}{1+z^{2^{k+1}}}=f(z)-\frac{z}{1+z}.\tag{2}$$ If we further assume: $$f(z)=\sum_{m\geq 1} a_m\,z^{m}\tag{3}$$ $(2)$ gives: $$a_k = a_{2k}+1,\qquad 0 = a_{2k+1}-1\tag{4}$$ hence:
$$f(z) = \sum_{m\geq 1}\left(1-\nu_2(m)\right) z^m\tag{5}$$
where $$\nu_2(m)=\max\{r\in\mathbb{N}:2^r\mid m\}.$$
• What function is $v_2$ representing here? – Brevan Ellefsen Sep 16 '15 at 20:56
• @BrevanEllefsen: the $2$-adic height, now written. – Jack D'Aurizio Sep 16 '15 at 20:57
• For short, the coefficient of $z^m$ in the Taylor expansion of $f(z)$ just depends on the number of trailing zeroes in the binary representation of $m$. – Jack D'Aurizio Sep 16 '15 at 20:59