If $\varphi_{k+1} \leq \varphi_k^2 \varphi_{k-1}$, then the sequence $(\varphi_k)_k$ converges of $R$-order $1+\sqrt2$ I am struggling with the following problem (I was thinking to apply Banach fixed-point theorem to the problem). Any help is appreciated.
Definition. Let $\left(\varepsilon_k\right)_{k \in \mathbb{N}_0} \subset \mathbb{R}^{+}$be a zero sequence, i.e. $\lim _{k \rightarrow \infty} \varepsilon_k=0$. We say the sequence converges (at least) with $\boldsymbol{R}$-order $p>1$ if there is a constant $\kappa \in(0,1)$ with
$$
\limsup _{k \rightarrow \infty} \frac{\varepsilon_k}{\kappa^{\left(p^k\right)}} \in \mathbb{R} .
$$
Given now a zero sequence $\left(\varphi_k\right)_{k \in \mathbb{N}_0} \subset \mathbb{R}^{+}$with.
$$
\varphi_{k+1} \leq \varphi_k^2 \varphi_{k-1}, \quad k \in \mathbb{N} .
$$
Prove that this sequence converges with $\boldsymbol{R}$-order $p=1+\sqrt{2}$!
 A: What we want to prove is basically the log-limit inequality version of the product version of a second order linear recurrence relation.

Since $\lim _{k \rightarrow \infty} \varphi_k=0$, there exist $c\in\Bbb N_0$ such that for all $n\ge c$, $\varphi_n<1$.
Let us construct a sequence for reference. Consider $(\psi_k)_{k\in\Bbb N_0, k\ge c}$ defined by $\psi_c=\ln(\varphi_c)<0$, $\psi_{c+1}=\ln(\varphi_{c+1})<0$, and
$$\psi_{k+1}=2\psi_k+\psi_{k-1}$$
for all $k\ge c+1$.
By straightforward induction, we can show that for all integer $k\ge c$,

*

*$\psi_k=a(1+\sqrt2)^{k-c}+b(1-\sqrt2)^{k-c}$, where $a=\frac{\psi_{c+1}-\psi_c(1-\sqrt2)}{2\sqrt2}<0$, $\ b=\frac{\psi_{c+1}-\psi_c(1+\sqrt2)}{-2\sqrt2}$.



*

*$\varphi_k\le e^{\psi_k}$
Since $|1-\sqrt2|<1$, $$\lim_{k\to\infty}(\psi_k-a(1+\sqrt2)^{k-c})=\lim_{k\to\infty}b(1-\sqrt2)^{k-c}=0.$$
Let $\kappa\in (e^{a(1+\sqrt2)^{-c}},1)$ and $p=1+\sqrt2$.
$$\begin{aligned}
\limsup _{k \rightarrow \infty} \frac{\varphi_k}{\kappa^{\left(p^k\right)}}
&\le\limsup _{k \rightarrow \infty} \frac{e^{\psi_k}}{\kappa^{\left(p^k\right)}}\\
&=\limsup _{k \rightarrow \infty} \frac{e^{a(1+\sqrt2)^{k-c}}}{\kappa^{\left(p^k\right)}}\\
&=\limsup _{k \rightarrow \infty} \frac{e^{(a(1+\sqrt2)^{-c})(1+\sqrt2)^{k}}}{\kappa^{\left(p^k\right)}}\\
&=\limsup _{k \rightarrow \infty} \left(\frac{e^{(a(1+\sqrt2)^{-c})}}\kappa\right)^{(1+\sqrt2)^{k}}\\ 
&=0
\end{aligned}$$
