Show that for \(|z|<1\), \[ (1+z+\cdots+z^9) (1+z^{10}+\cdots+z^{90}) (1+z^{100}+\cdots+z^{900})\cdots =\frac1{1-z}. \]
Solution
For \(j\geq0\), define \[ F_j(z) \coloneqq \sum_{d=0}^{9}z^{d10^j}. \]
1. For every \(j\geq0\) and every \(z\) with \(z^{10^j}\neq1\),\[ F_j(z) = \frac{1-z^{10^{j+1}}}{1-z^{10^j}}. \]
Apply the finite geometric-series formula \[ 1+w+\cdots+w^9 = \frac{1-w^{10}}{1-w} \] with \[ w=z^{10^j}. \] Then \[ w^{10}=z^{10^{j+1}}, \] which gives the stated identity.
2. If \(\abs{z}<1\), then for every \(N\geq0\),\[ \prod_{j=0}^{N}F_j(z) = \frac{1-z^{10^{N+1}}}{1-z}. \]
When \(\abs z<1\), none of the numbers \(z^{10^j}\) equals \(1\), so step 1 applies to every factor. Thus \[ \begin{aligned} \prod_{j=0}^{N}F_j(z) &= \prod_{j=0}^{N} \frac{1-z^{10^{j+1}}}{1-z^{10^j}}\\ &= \frac{1-z^{10^{N+1}}}{1-z}, \end{aligned} \] because all intermediate factors cancel.
3. If \(\abs{z}<1\), then\[ z^{10^{N+1}}\longrightarrow0 \qquad(N\to\infty). \]
Since \(0\leq\abs z<1\), \[ \abs{z^{10^{N+1}}} = \abs z^{10^{N+1}} \longrightarrow0. \]
4. Therefore, for \(\abs z<1\),\[ \boxed{ \prod_{j=0}^{\infty} \left( \sum_{d=0}^{9}z^{d10^j} \right) = \frac1{1-z} }. \]
The infinite product is the limit of its finite partial products. By step 2, \[ \prod_{j=0}^{N}F_j(z) = \frac{1-z^{10^{N+1}}}{1-z}, \] and step 3 makes the numerator tend to \(1\). Hence the limit is \(1/(1-z)\).
5. Q.E.D.
Step 4 is exactly the claimed product identity.