Prove that $ \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z} $

Show that for $|z|<1$, one have $$ \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots+\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots = \frac{z}{1-z} $$ and $$ \frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots+\frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots = \frac{z}{1-z} $$ Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots+2^k$.

Solution 1:

HINT: Note that

$$\begin{align*} \frac{z^{2^k}}{1-z^{2^{k+1}}}&=z^{2^k}\sum_{n\ge 0}\left(z^{2^{k+1}}\right)^n\\ &=z^{2^k}\sum_{n\ge 0}z^{2^{k+1}n}\\ &=\sum\left\{z^n:1\le n\equiv 2^k\pmod{2^{k+1}}\right\}\;, \end{align*}$$

so

$$\begin{align*} \sum_{k\ge 0}\frac{z^{2^k}}{1-z^{2^{k+1}}}&=\sum_{k\ge 0}\sum\left\{z^n:1\le n\equiv 2^k\pmod{2^{k+1}}\right\}\\ &=\sum\bigcup_{k\ge 0}\left\{z^n:1\le n\equiv 2^k\pmod{2^{k+1}}\right\}\\ &=\sum\left\{z^n:\exists k\ge 0\left(n\equiv 2^k\pmod{2^{k+1}}\right)\right\}\;. \end{align*}$$

Moreover,

$$\frac{z}{1-z}=\sum_{n\ge 1}z^n\;.$$

Now show that for each positive integer $n$ there is a unique non-negative integer $k$ such that $n\equiv 2^k\pmod{2^{k+1}}$. The dyadic expansion of $n$ may help you here.

For the second question, note that

$$\frac{2^kz^{2^k}}{1+z^{2^k}}=2^kz^{2^k}\sum_{n\ge 0}(-1)^nz^{2^kn}=2^k\sum_{n\ge 0}(-1)^nz^{2^k(n+1)}\;,$$

so

$$\sum_{k\ge 0}\frac{2^kz^{2^k}}{1+z^{2^k}}=\sum_{k\ge 0}2^k\sum_{n\ge 0}(-1)^nz^{2^{k(n+1)}}\;.\tag{1}$$

Show that if if $n=2^km$, where $k\ge 0$ and $m$ is odd, then the coefficient of $z^n$ in $(1)$ is $$\sum_{\ell=0}^k(-1)^{2^{k-\ell}m-1}2^\ell\;,$$ and evaluate this finite sum to find the coefficient of $z^n$ in $(1)$.