Find $\prod^\infty_{n=1} (1 - 10^{-3* 2^n})$.

Question

Find the infinite product: $0.999 \times 0.999999 \times 0.999999999999 \cdots$, where the number of $9$s in each term is double that of the previous term.

I am thinking of considering the analogous problem $0.9 \times 0.99 \times 0.999 \cdots$, but this alone seems quite hard to solve. To solve this analogous problem, I was thinking of considering the numbers that have a binary sum of digits of $1$, $2$, etc., but this does not give anything nice.

On the other hand, is there a nice way to express

$\frac{0.999 \times 0.999999 \times 0.999999999999 \cdots}{0.9 \times 0.99 \times 0.999 \cdots}$?

Answer 1

It seems that you are looking for $$P_k=\prod _{n=0}^k \left(1-10^{-3\times 2^{n}}\right)$$ I must confess that I do not see any possible closed form but we can notice very interesting patterns as shown below $$\left( \begin{array}{cc} k & P_k \\ 0 & 0.99900000000000000000000000000000000000000000000000000000000000000 \\ 1 & 0.99899900100000000000000000000000000000000000000000000000000000000 \\ 2 & 0.99899900099900100099900000000000000000000000000000000000000000000 \\ 3 & 0.99899900099900100099899900100099900099899900100000000000000000000 \\ 4 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 5 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 6 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 7 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 8 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 9 & 0.99899900099900100099899900100099900099899900099900100099900099900 \\ 10 & 0.99899900099900100099899900100099900099899900099900100099900099900 \end{array} \right)$$

Answer 2

We can generalise this problem as follows: $$s = \prod_{n =0}^{\infty}(1-x^{2^n}) = (1-x)(1-x^2)(1-x^4)(1-x^8) ...$$ where $x=0.001$ in your specific problem.

If we expand this product as $$s=1+\sum_{m=1}^{\infty}a_mx^m$$ then each $a_m$ is either 1 or -1 because there is only one way to express each m as the sum of powers of two (binary expansion). And we can go further and see that $a_m=(-1)^{t_m}$ where $t_m$ is the number of 1s in the binary expansion of m ($t_m$ is related to the Thue-Morse sequence). If we say that $t_0=0$ then we have $$s=\sum_{m=0}^{\infty}(-1)^{t_m}x^m=1-x-x^2+x^3-x^4+x^5+x^6-x^7-x^8 ...$$ I don't know of a simpler way of expressing this limit.