Digits in $2^n$

This problem was heard from here. Let $a_n$ be the number of odd digits in the base-$10$ expansion of $2^n$. Prove that

$$
\sum_{n=1}^{\infty}\frac{a_n}{2^n}=\frac{1}{9}.
$$

What a amazing property!

=====================

Proof:

Write $2^n$ in decimal digits $\ldots d_{n,1}d_{n,0}$. Observe that $d_{i,j}$ is odd if and only if $5\leq d_{i-1,j-1}\leq 9$, which happens if and only if $d_{i-2,j-1}d_{i-2,j-2}\in \{[25,49], [75,99]\}$, which....and so on. So

Claim 1

$d_{n,i}$ is odd if and only if $10^{-i}\in\cup_{k=1}^{2^n}[k2^{-n}-2^{-n-1},k2^{-n})$.

Immediately we have

Claim 2

$d_{n,i}$ is odd if and only if the $n$-th digit after decimal point in $10^{-i}$'s binary representation is $1$.

Now we have everything. The sum $\sum_{n=1}^{\infty}\frac{a_n}{2^n}$ is nothing but $\sum_{i=1}^{\infty}10^{-i}$, which is $\frac{1}{9}$.

Q.E.D.