positivity of recursion $x_{n+2}=x_{n+1}-\rho ^2 x_n$ $M_n =I_n +\rho (J_{n-1}+J_{n-1}^T)$, let $x_n=\det M_n$.
Asking when the terms of $x_{n+2}=x_{n+1}-\rho ^2 x_n, \\x_0=x_1=1$ are positive for $n>0$, is equivalent to asking when the related matrices are all positive-definite.
Choose vector 
$v=\left( v_i \right), v_i=(-1)^i$
and calculate $v^TM_nv$, we conclude that $\rho \leq \frac12$.
Is there a direct approach to prove that if $\rho>\frac12$, then $\{x_n\}_{n\ge 0}$ are not positive? If I do not calculate wrongly, it asks to show that the inequalities
$$\sum_{k=0}^{\left[\frac{n-1}4\right]}{n\choose {4k+1}}(2b)^{4k}>
\sum_{k=0}^{\left[\frac{n-3}4\right]}{n\choose {4k+3}}(2b)^{4k+2}
$$
where $2b=\sqrt{4\rho^2-1}$,
do not hold for all $n>1$ if $b>0$.
 A: 
Is there a direct approach to prove that if $\rho>\frac12$, then $\{x_n\}_{n\ge 0}$ are not positive?

$x_{n+2} - x_{n+1} + \rho^2 x_n = 0\,$ is a second-order linear homogeneous recurrence with constant coefficients. Its characteristic polynomial $\,t^2-t+\rho^2\,$ has complex roots iff $\,\rho \gt \frac{1}{2}\,$. In that case, since the roots $\,t_1=\overline{t_2}\,$ are complex conjugates and $\,t_1t_2 = \rho^2\,$, it follows that $\,|t_1|=|t_2|=\rho\,$, so the roots can be written as $\,t_{1,2}=\frac{1}{2}\left(1 \pm i \sqrt{4 \rho^2-1}\right)=\rho e^{\pm i \alpha}\,$ where $\,0 \lt \alpha \lt \pi\,$ (with strict inequalities, since $t_{1,2}$ are not real).
The general solution of the recurrence is then $\,x_n=c_1\,t_1^n+c_2\,t_2^n=c_1\,t_1^n+c_2\,{\overline{t_1}}{\,^n}\,$. Since the initial values $\,\,x_{1,2}$ are real, it follows that all terms of the sequence are real, which implies $\,c_2 = \overline{c_1}\,$, so the general solution can be written as $\,x_n = c_1 t_1^n + \overline{c_1} \overline{t_1}^{\,n} = 2 \operatorname{Re}(c_1 t_1^n)\,$.
Let $\,c_1 = c e^{i \gamma}\,$ in polar form with real $\,c \gt 0\,$ and $\,0 \le \gamma \lt 2 \pi\,$, then in the end: $$x_n = 2 \operatorname{Re}\left(c \rho^n e^{i(\gamma + n \alpha)}\right)=2c\rho^n \cos(\gamma + n \alpha)$$
Since $\,0 \lt \alpha \lt \pi\,$, there will be at least one term of the form $\,\gamma + n \alpha\,$ falling into any interval $\,(a,b)\,$ of length $\,b-a\ge \pi\,$ starting past $\,a \ge \gamma + \alpha\,$. In particular, there will be at least one $\,n\,$ such that $\,2k\pi + 4\pi + \frac{\pi}{2} \lt \gamma + n \alpha \lt 2k\pi + 4 \pi + \frac{3 \pi}{2}\,$ for each integer $\,k \ge 0\,$, and for all those values of $\,n\,$ the factor $\,\cos(\gamma + n \alpha) \lt 0\,$ is negative, therefore $\,x_n = 2c\rho^n \cos(\gamma + n \alpha) \lt 0\,$.
