Boundedness of a real sequence Let $\{x_k\}_{k=1}^\infty$ be a sequence of real numbers. Let $\theta\in(0,\pi/2)$, and assume that for all $k\in\mathbb{Z}^+$, 
\begin{equation}
|x_{2k}-e^{\jmath\theta}x_{2k+1}|^2-|x_{2k-1}-e^{\jmath\theta}x_{2k}|^2\leq 0.
\end{equation}
Can I prove that $x_k$ is a bounded sequence? If not what's a disproof(counterexample)? Thanks
Note: $\mathbb{Z}^+$ denotes the set of all positive integers and $\jmath$ denotes $\sqrt{-1}.$
 A: Note that $$\lvert x_{2k}-e^{i\theta}x_{2k+1}\rvert^2 = x_{2k}^2-2\cos(\theta)x_{2k}x_{2k+1}+x_{2k+1}^2$$ and $$\lvert x_{2k-1}-e^{i\theta}x_{2k}\rvert^2 = x_{2k-1}^2-2\cos(\theta)x_{2k-1}x_{2k}+x_{2k}^2$$ Therefore, $$\lvert x_{2k}-e^{i\theta}x_{2k+1}\rvert^2-\lvert x_{2k-1}-e^{i\theta}x_{2k}\rvert^2 = x_{2k+1}^2-2\cos(\theta)(x_{2k}x_{2k+1}-x_{2k-1}x_{2k})-x_{2k-1}^2 = (x_{2k+1}-x_{2k-1})(x_{2k+1}+x_{2k-1}-2\cos(\theta)x_{2k})$$ Considering that $\cos(\theta) > 0$, this is less than or equal to $0$ iff $x_{2k+1}\geq x_{2k-1}$ and $x_{2k}\geq \frac{x_{2k+1}+x_{2k-1}}{2\cos(\theta)}$ or $x_{2k+1}\leq x_{2k-1}$ and $x_{2k}\leq \frac {x_{2k+1}+x_{2k-1}}{2\cos(\theta)}$. One specific counterexample could be $\theta = \frac{\pi}{3}$ and $$x_n = \begin{cases} n,& n\text{ odd} \\ 2n+1,& n\text{ even}\end{cases}$$ which is unbounded.
A: Just take $x_k=k$ and let $\jmath=0$.
Update: some need more explanation than others, take $\theta=0$ before the post was edited.
A: Here's me thinking out-loud so to speak:
Well $e^{\jmath\theta}$ just rotates $x_{2k+1}$ in the first term, and likewise in the second.  So $|x_{2k-1}-e^{\jmath\theta x_{2k}}|$ represents the distance between $x_{2k-1}$ and the rotation of $x_{2k}$ by an angle $\theta$.  This distance will be greater-than-or-equal-to the distance between $x_{2k}$ and the rotation by $\theta$ of $x_{2k+1}$. In effect we want a sequence in which, if you rotate adjacent terms, the distances due to rotations does not grow.
If we wanted to try to find a counterexample we might look for an unbounded sequence which satisfies this.  We might guess $x_k=k$.  But here the distance between terms and the rotation of subsequent terms grows for all $\theta$ except $\theta=0$.  But apparently some cruel deity has forbidden the choice $\theta = 0$.  So be it, we can then rotate our sequence terms one after the other, but preserve roughly the same answer.  If we pick $\theta=\pi/4$ then 
$x_k = ke^{\jmath k\pi/4}$
