Two different expansions of $\frac{z}{1-z}$

Solution 1:

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.

Solution 2:

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: $$...=\sum_{k=0}\left(\sum_{t\in A_k}z^t\right)=\sum_{t\in A_0}z^t+ \sum_{k=1}\left(\sum_{t\in A_k}z^t\right)=\sum_{s=0}z^{2s+1} + \sum_{k=1}\left(\sum_{s=0}z^{2^k(2s+1)}\right)=\\ z\sum_{s=0}z^{2s}+\sum_{k=1}z^{2^k}\left(\sum_{s=0}z^{2^k(2s)}\right)=\frac{z}{1-z^2}+\sum_{k=1}\frac{z^{2^k}}{1-z^{2^{k+1}}}$$