I would like to study the sequence defined by $$ a_n=d(u_n,\Bbb{Z}). $$
Where $u_n=\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n$ and $d(u_n,\Bbb{Z})=\inf\{d(u_n,x):x\in\Bbb{Z}\}$.
I do not really have ideas, here is some 'though',
$$a_{n+1}-a_n=d(u_{n+1},\Bbb{Z})-d(u_n,\Bbb{Z})=\inf(\{d(u_{n+1},x):x\in\Bbb{Z}\})-\inf(\{d(u_{n},x):x\in\Bbb{Z}\}).$$
Now by theorem I have $\vert d(u_{n+1},\Bbb{Z})-d(u_n,\Bbb{Z})\vert\le\Vert u_{n+1}-u_n\Vert$.
Let us compute $\Vert u_{n+1}-u_n\Vert=\Vert\bigl(\frac{1+\sqrt{5}}{2}\bigr)^{n+1}-\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n\Vert=\Vert\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n\frac{\sqrt{5}-1}2\Vert.$
After that I am not sure what I have to do. Any ideas?
Hint: Show that $$\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n$$ is an integer. The second term has limit $0$.
Remark: There are many ways to show that our expression $\alpha^n+\beta^n$ is an integer. One way is by induction, using the fact that $$\alpha^{n+1}+\beta^{n+1}=(\alpha+\beta)(\alpha^n+\beta^n)-\alpha\beta(\alpha^{n-1}+\beta^{n-1}),$$ and $\alpha+\beta$ and $\alpha\beta$ are integers.
Another way is to show that each of $\alpha^n$ and $\beta^n$, and therefore their sum, satisfies the recurrence $x_{n+1}=x_n+x_{n-1}$. Then from the fact that $\alpha^0+\beta^0$ and $\alpha^1+\beta^1$ are integers, we can conclude that $\alpha^n+\beta^n$ is an integer for all $n$. By the way, these are the Lucas numbers, close relatives of the Fibonacci numbers.
Whenever $\frac{1+\sqrt{5}}{2}$ has a problem, its conjugate $\frac{1-\sqrt{5}}{2}$ is ready to pitch in and help.