Show that for \(|z|<1\), \[ \frac{z}{1-z^2} +\frac{z^2}{1-z^4} +\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}} +\cdots =\frac{z}{1-z}, \] and \[ \frac{z}{1+z} +\frac{2z^2}{1+z^2} +\cdots +\frac{2^kz^{2^k}}{1+z^{2^k}} +\cdots =\frac{z}{1-z}. \] Justify any change in the order of summation.
Solution
1. For every \(k\ge0\) and \(|z|<1\),\[ \frac{z^{2^k}}{1-z^{2^{k+1}}} = \sum_{j=0}^{\infty}z^{2^k(2j+1)}. \]
Since \(|z^{2^{k+1}}|<1\), the geometric-series formula gives \[ \frac{1}{1-z^{2^{k+1}}} = \sum_{j=0}^{\infty}z^{j2^{k+1}}. \] Multiplying by \(z^{2^k}\) yields the stated expansion.
2. Every positive integer \(n\) can be written uniquely in the form\[ n=2^k(2j+1), \qquad k,j\in\mathbb Z_{\ge0}. \]
Take \(k\) to be the largest exponent such that \(2^k\) divides \(n\). Then \(n/2^k\) is odd, hence equals \(2j+1\) for a unique \(j\ge0\). Maximality of \(k\) gives uniqueness.
3. The double series obtained from 1 is absolutely convergent for \(|z|<1\).
Put \(r=|z|<1\). By 2, the exponents \(2^k(2j+1)\), as \((k,j)\) ranges over nonnegative integers, run through every positive integer exactly once. Therefore \[ \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} \left|z^{2^k(2j+1)}\right| = \sum_{n=1}^{\infty}r^n = \frac{r}{1-r}<\infty. \]
4. Hence\[ \sum_{k=0}^{\infty}\frac{z^{2^k}}{1-z^{2^{k+1}}} = \frac{z}{1-z}. \]
By 1 and absolute convergence from 3, the double series may be rearranged. Using the bijection in 2, \[ \sum_{k=0}^{\infty}\sum_{j=0}^{\infty}z^{2^k(2j+1)} = \sum_{n=1}^{\infty}z^n = \frac{z}{1-z}. \]
5. For every \(k\ge0\) and \(|z|<1\),\[ \frac{2^kz^{2^k}}{1+z^{2^k}} = \frac{2^kz^{2^k}}{1-z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}. \]
For any \(x\ne\pm1\), \[ \frac{x}{1+x} = \frac{x}{1-x}-\frac{2x^2}{1-x^2}. \] Apply this with \(x=z^{2^k}\) and multiply by \(2^k\).
6. The \(N\)th partial sum of the second series is\[ \sum_{k=0}^{N}\frac{2^kz^{2^k}}{1+z^{2^k}} = \frac{z}{1-z} - \frac{2^{N+1}z^{2^{N+1}}}{1-z^{2^{N+1}}}. \]
Set \[ A_k=\frac{2^kz^{2^k}}{1-z^{2^k}}. \] By 5, the \(k\)th summand is \(A_k-A_{k+1}\). The finite sum therefore telescopes to \(A_0-A_{N+1}\), and \(A_0=z/(1-z)\).
7. For \(|z|<1\),\[ \frac{2^{N+1}z^{2^{N+1}}}{1-z^{2^{N+1}}}\longrightarrow0. \]
Let \(r=|z|<1\). For all sufficiently large \(N\), \(r^{2^{N+1}}\le1/2\), so \[ \left|\frac{2^{N+1}z^{2^{N+1}}}{1-z^{2^{N+1}}}\right| \le 2^{N+2}r^{2^{N+1}}. \] Writing \(c=-\log r>0\), the right-hand side is \[ 2^{N+2}e^{-c2^{N+1}}, \] which tends to \(0\) because the exponential decay in \(2^N\) dominates the factor \(2^N\).
8. Therefore\[ \sum_{k=0}^{\infty}\frac{2^kz^{2^k}}{1+z^{2^k}} = \frac{z}{1-z}. \]
Let \(N\to\infty\) in the finite telescoping identity of 6 and use 7.