- Prove that the sequence {$\gamma_{n}$}$_{n=1}^{\infty},$ where $\gamma_{n}$ is the fractional part of $$\bigl( \frac{1+\sqrt{5}}{2}\bigr)^{n},$$ is not equidistributed in $[0,1].$
[Hint: Show that $U_{n}=\bigl( \frac{1+\sqrt{5}}{2}\bigr)^{n}+\bigl( \frac{1-\sqrt{5}}{2}\bigr)^{n}$ is the solution of the difference equation $U_{r+1}=U_{r}+U_{r-1}$ with $U_{0}=2$ and $U_{1}=1.$ The $U_{n}$ satisfy the same difference equation as the Fibonacci numbers.]
I got the answer from this site :
Proof. According to the hint, one observes that $U_{n}= W_{n}+\bar W_{n}$ satisfies $U_{r}=U_{r-1}+U_{r-2}$ for $r \ge 2$ and $U_{0}=2$ and $U_{1}=1.$ So $U_{n}\in \Bbb N$ for all $n\in \Bbb Z_{\ge 0}.$ Note that $\bar{W}_{n}\to 0$ as $n\to \infty$ since $\vert \frac{1-\sqrt{5}}{2}\vert<1.$ So we complete the proof since $\gamma_{n}\notin (\frac{1}{4},\frac{3}{4})$ for all large $n$.
I can't understand why particularly $\gamma_{n}\notin (\frac{1}{4},\frac{3}{4})$.
$\overline W_n\to0$ as $n\to\infty$ means that, for sufficiently large $n, |\overline W_n|<\frac14.$
(In fact, that would be true for any positive $\epsilon$ in place of $\frac14.)$
If $|\overline W_n|=|U_n-W_n|<\frac14<1$ and $U_n$ is an integer, then
if $W_n>U_n$ then $\gamma_n=W_n-U_n<\frac14$
and if $W_n\lt U_n$ then $\gamma_n=W_n+1-U_n=1-\overline W_n>\frac{3}4,$
so $\gamma_n \not\in(\frac14,\frac34)$.