Complex series sum

Question

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.

Answer 1

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?

Answer 2

$$\begin{align} \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} &= \sum_{n=0}^\infty z^{2^n}\sum_{k=0}^\infty z^{k2^{n+1}} \\&= \sum_{n=0}^\infty \sum_{k=0}^\infty z^{2^n+k2^{n+1}} \\&= \sum_{n=0}^\infty \sum_{k=0}^\infty z^{2^n(1+2k)} \end{align}$$ 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 $i\geqslant0$.

Answer 3

"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 $2$".