Complex series sum Show that for $|z|<1$, 
$$\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z}$$
and
$$\sum_{n=0}^\infty \frac{2^n z^{2^n}}{1+z^{2^n}}=\frac{z}{1-z}$$
As a hint is given to use the dyadic expansion of an integer.
I have no idea how to proceed.Please help.
 A: The LHS of the first is
\begin{align}
&(z+z^3+z^5+z^7+\cdots)\\
&+(z^2+z^6+z^{10}+z^{14}+\cdots)\\
&+(z^4+z^{12}+z^{20}+z^{28}+\cdots)+\cdots.
\end{align}
The LHS of the second is
\begin{align}
&(z-z^2+z^3-z^4+z^5-\cdots)\\
&+(2z^2-2z^4+2z^6-2z^8+2z^{10}-\cdots)\\
&+(4z^4-4z^8+4z^{12}-4z^{16}+4z^{20}-\cdots)+\cdots.
\end{align}
Do these give a clue?
A: $$\begin{align}
\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} &= \sum_{n=0}^\infty z^{2^n}\sum_{k=0}^\infty z^{k2^{n+1}} \\&=
\sum_{n=0}^\infty \sum_{k=0}^\infty z^{2^n+k2^{n+1}} \\&=
\sum_{n=0}^\infty \sum_{k=0}^\infty z^{2^n(1+2k)}
\end{align}$$
Now apply your hint, which can be worded as

Every integer $k\geqslant1$ can be written in one and only one way as $k=2^n+i2^{n+1}=2^n(1+2i)$, for some $n\geqslant0$ and $i\geqslant0$.

A: "Any positive integer has a unique representation in base $2$" leads to
$$ \prod_{k\geq 0}\left(1+z^{2^k}\right) = 1+z+z^2+z^3+\ldots = \frac{1}{1-z} \tag{A}$$
and by applying $\frac{d}{dz}\log(\cdot)$ to both sides we get:
$$ \sum_{k\geq 0} \frac{2^k z^{2^k-1}}{z^{2^k}+1} = \frac{1}{1-z}\tag{B}$$
proving your second identity. The first identity has already been proved by Brevan Ellefsen: it is a consequence of "Any positive integer can be written in a unique way as the product between an odd integer and a power of $2$".
