TITLE: Complex series sum
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: "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 $2$".
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