Periodicity and floor functions in 2016!

In general, we use base 10 to represent any real number. For example $652.333333...=6*10^{2}+5*10^{1}+2+\frac{3}{10}+\frac{3}{10^{2}}+\frac{3}{10^{3}}+....=652\frac{1}{3}.$ So if $x$ is any real number then there is sequence $a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n}, ...,$ such that any $a_{i}$ is a digit, and $x=\sum_{k=0}^{\infty}a_{k}*10^{m-k}$ for a certain integer $m$ that depends on $x$ . In particular, the number $x=\dfrac{10^{2015}-1}{10^{2016}-1}$ can be represented by a series $x=\sum_{k=0}^{\infty}b_{k}*10^{-k}.$ Now it is easy to see that for any integer $n\geq 0,$ $f(n)= \left \lfloor 10^{n+1}x \right \rfloor - 10\left \lfloor 10^n x \right \rfloor=b_{n+1}.$ Since $x$ is a rational number, its decimal representation is periodic. Therefore $f(n)$ is a periodic function. By direct calculations, it can be proved that the period of $f$ is 2016.