Let $a(x)$ be the sequence defined as $$a_1(x)=\frac x2\\a_{n+1}(x)=\frac{20a_n(x)-19\left\lfloor a_n(x)\right\rfloor}{2}$$ for all $x,n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{\left\lfloor a_k(x)\right\rfloor}{10^k}=\frac{x}{19}$$
The first thing I have tried is to group terms of given sum. However, it is very hard because of floor function inside it. My second attempt is to use induction because $x$ is natural number. After many attempts, I could'n prove it using this approach.
Then I decided to expand $\frac x{19}$ in some way and try to get above sum from that. I wrote $\frac x{19}$ as
$$\frac{x}{19}=\frac{\frac x{20}}{1-\frac1{20}}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{x}{20^k}$$
Now I only need to prove that this sum is equal to above sum. I have tried to group and cancel out some terms, but I could get rid of floor function. Am I missing something obvious?