TITLE: Recurrence relations and limits, tough.
QUESTION [19 upvotes]: I would like a hint for the following, more specifically, what strategy or approach should I take to prove the following?
Problem: Let $P \geq 2$ be an integer. Define the recurrence
$$p_n = p_{n-1} + \left\lfloor \frac{p_{n-4}}{2} \right\rfloor$$
with initial conditions:
$$p_0 = P + \left\lfloor \frac{P}{2} \right\rfloor$$
$$p_1 = P + 2\left\lfloor \frac{P}{2} \right\rfloor$$
$$p_2 = P + 3\left\lfloor \frac{P}{2} \right\rfloor$$
$$p_3 = P + 4\left\lfloor \frac{P}{2} \right\rfloor$$
Prove that the following limit converges:
$$\lim_{n\rightarrow \infty} \frac{p_n}{z^n}$$
where $z$ is the positive real solution to the equation $x^4 - x^3 - \frac{1}{2} = 0$.
Note: I've already proven the following:
$$\lim_{n\rightarrow \infty} \frac{p_n}{p_{n-1}} = z$$
Any ideas? Not sure if this result helps. Also $\lim_{n\rightarrow \infty}p_n/z^n$ is also bounded above and below. I've attempted to show $\lim_{n\rightarrow \infty} \frac{p_n}{z^n}$ is Cauchy, but had no luck with that. I don't know what the limit converges to either.
Edit: I believe the limit should converge as $p_n$ achieves an end behaviour of the form $cz^n$ for $c \in \mathbb{R}$ (this comes from the fact that the limit of the ratios of $p_n$ converge to $z$), however I do not know how to make this rigorous.
Edit 2: Proving the limit exists is equivalent to showing
$$p_0 \cdot \prod_{n=1}^{\infty} \left( \frac{p_n/p_{n-1}}{z} \right)$$
converges.
UPDATED:
If someone could prove that $|p_n-z \cdot p_{n-1}|$ is bounded above (or converges, or diverges), then the proof is complete.

REPLY [2 votes]: Let us start with the solution of the homogeneous recurrence 
$$\phi_n = \phi_{n-1} + \frac{\phi_{n-4}}{2}$$
its characteristic equation is 
$$x^4 - x^3 - \frac{1}{2} = 0$$
this equations has $4$ solutions, two of them are complex and the other two are a negative and a positive real number. Their approximate values, as given by Mathematica, are: 
$$z_1=1.25372, \ \ z_2=-0.669107, \ \ z_3=0.207691 + 0.743573 i, \ \ z_4=0.207691 - 0.743573 i,$$
(in your answer you label $z$ the one labeled $z_1$ above). Notice that the positive real solution $z=z_1=1.25372$ is the one with the greatest magnitude among the $4$ solutions (actually, it is the only one whose magnitude exceeds $1$). 
Now, the general solution to the homogeneous recurrence is:
$$\phi_n=c_1z^n_1+c_2z^n_2+c_3z^n_3+c_4z^n_4$$ 
where $c_1, c_2, c_3, c_4$ are constants to be determined from the initial conditions posed in your question. Since $z=z_1=1.25372$ has the greatest magnitude among the roots of the characteristic equation, the above general  solution asymptotically (for $n$ large enough) tends to $c_1z^n$ i.e.
$$\phi_n\sim c_1z^n$$
Consequently, 
$$\frac{\phi_n}{z^n}\sim c_1 \ \ \textrm{ i.e. } \ \  \lim_{n\rightarrow\infty}\frac{\phi_n}{z^n}=c_1$$
where $c_1$ will be determined by the solution of the $4\times 4$ linear system of equations 
$$
\phi_i=c_1z^i_1+c_2z^i_2+c_3z^i_3+c_4z^i_4
$$
for $i=0,1,2,3$, $p_i$ given by the initial conditions posted in the question and $z_1=z,z_2,z_3,z_4$ the roots of the characteristic equation given above. 
Let me now try to justify why the convergence of $\frac{\phi_n}{z^n}$ implies also the convergence of $\frac{p_n}{z^n}$. The recurrence 
$$p_n = p_{n-1} + \left\lfloor \frac{p_{n-4}}{2} \right\rfloor = p_{n-1} + \frac{p_{n-4}}{2} - \epsilon_n$$
differs from the homogeneous, by a bounded function $0\leq\epsilon_n< 1$ of $n$. Since we are dealing with linear recurrences, increasing sequences $p_n$, $\phi_n$ and we are interested in the asymptotic behaviour of the solutions, in the limit of large $n$, the two are essentially the same. The solutions $p_n$ and $\phi_n$ differ by a $O(1)$ special solution (of the posted, non-homogeneous  recurrence): 
$$
p_n=\phi_n+O(1) \Rightarrow p_n\sim\phi_n\Rightarrow\frac{p_n}{z^n}\sim \frac{\phi_n}{z^n}\Rightarrow\lim_{n\rightarrow\infty}\frac{p_n}{z^n}=c_1
$$
 We can also see that the bigger the value of $P\geq 2$ (given in the initial conditions), the quicker  $\frac{p_n}{z^n}$ converges.
P.S.  Regarding the estimation that the general solutions $p_n$ and $ϕ_n$ of the respective recurrences, differ by a $O(1)$ special solution of the non-homogeneous, my argument is the following: when dealing with non-homogeneous linear recurrences with constant coefficients i.e.
$$
p_n+c_1p_{n−1}+...+c_dp_{n−d}=h(n)
$$
and $h(n)=const$, then it is customary to seek for a special solution to be a constant. Since here, the non-homogeneous factor is the bounded function $0\leq\epsilon_n< 1$, I guess that it is reasonable to conjecture that the corresponding special solution is $O(1)$.