Two different expansions of $\frac{z}{1-z}$ This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.
Show that for $|z|<1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and 
$$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.$$
Justify any change in the order of summation. 
[Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots +2^k$.]
I don't really know how to work this through. I know that $\frac{z}{1-z}=\sum_{n=1}^\infty z^n$ and each $n$ can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated.
 A: Since minimalrho has explained how to proceed with the given hint, I'll give an alternative method. The $k$th summand of the first series can be written 
$$\frac{z^{2^k}}{1 - z^{2^{k}}} - \frac{z^{2^{k+1}}}{1-z^{2^{k+1}}}$$
and the $k$th summand of the second series can be written
$$\frac{2^kz^{2^k}}{1 - z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}$$
Hence, the $N$th partial sums of the two series telescope to 
$$\frac{z}{1 - z} - \frac{z^{2^{N+1}}}{1 - z^{2^{N+1}}}\quad \text{and}\quad \frac{z}{1 - z} - \frac{2^{N+1}z^{2^{N+1}}}{1 - z^{2^{N+1}}}$$
respectively. Using the condition $\lvert z\rvert < 1$, argue that $z^{2^{N+1}}/(1 - z^{2^{N+1}})$ and $2^{N+1}z^{2^{N+1}}/(1 - z^{2^{N+1}})$ tend to $0$ as $N\to \infty$. Then the results follow.
A: Hint and partial answer: Using this partition of integers and
$$\frac{z}{1-z}=\sum_{n=1}z^n=...$$
This series is absolute converging given $|z|<1$, thus changing the order of summation doesn't affect the final value. As a result:
$$...=\sum_{k=0}\left(\sum_{t\in A_k}z^t\right)=\sum_{t\in A_0}z^t+ \sum_{k=1}\left(\sum_{t\in A_k}z^t\right)=\sum_{s=0}z^{2s+1} + \sum_{k=1}\left(\sum_{s=0}z^{2^k(2s+1)}\right)=\\
z\sum_{s=0}z^{2s}+\sum_{k=1}z^{2^k}\left(\sum_{s=0}z^{2^k(2s)}\right)=\frac{z}{1-z^2}+\sum_{k=1}\frac{z^{2^k}}{1-z^{2^{k+1}}}$$
A: Hint for the first sum. Note that each positive integer $n$ can be written in a unique way as the product of a power of $2$, $2^k$, and an odd number $(2j+1)$.
Hint for the second sum. Note that if $n=2^k(2j+1)$ then the coefficient of $z^n$ of the left-hand side is
$$-1-2-2^2-\cdots -2^{k-1}+2^{k}.$$
