Infinite product $\prod\limits_{k=0}^\infty\sum\limits_{n=0}^9z^{10^kn} $ leading to $1/(1-z)$

Question

Please give me a hint (i am studying Complex Variables for Engineering) on how to prove that

$(1+z+z^2+\cdots+z^9)(1+z^{10}+z^{20}+\cdots+z^{90})\cdots=\prod\limits_{k=0}^\infty\sum\limits_{n=0}^9z^{10^kn}=\dfrac1{1-z}$

for complex $z:|z|<1$

I guess that it involves the Geometric Series (formula given modulus of $z$), but i want to have a precise start.

Answer 1

Hint. You may observe that $$ \sum\limits_{n=0}^9z^{10^kn}=\frac{1-z^{10^{k+1}}}{1-z^{10^k}} $$ then recognize a telescoping product: $$ \prod\limits_{k=0}^N\sum\limits_{n=0}^9z^{10^kn}=\prod\limits_{k=0}^N\frac{1-z^{10^{k+1}}}{1-z^{10^k}}. $$ Letting $N \to \infty$ gives the result.

Answer 2

When you multiply a polynomial, you take every permutation of monomials from each product. As a familiar example,

$$(x+1)(x+2) = x^2+x+2x+1 \cdot 2$$

In the case given, when considering a monomial in the final product, the first factor decides the one's place, the second factor decides the hundred's place, etc. Since the numbers $0-9$ can be used for each place as seen in the product given, every combination of digits will be made exactly once, which multiplies out to $$1+x+x^2+x^3+...$$

Which is equivalent to what you have by geometric series.

This method of multiplication can allow you to manipulate the expression; for example, removing the $z^{20}$ in the second factor will give you

$$1+x+...+x^{19}+x^{30}+...+x^{119}+x^{130}+...$$