Upper bound of a sequence for all n Let $r_n$ is a non-increasing sequence such that $ 1 \leq r_n \leq 1 + \theta_1^2 + \theta_2^2$ and $r_n \rightarrow 1$ as $n \rightarrow \infty$. Further we know that $|\theta_2| < 1$ and $|\theta_1| < 1 + \theta_2$.
If
\begin{equation}
\theta_{n, 1} = \frac{\theta_1(1+\theta_2) - \theta_2\theta_{n-1, 1}}{r_n}
\end{equation}
how can I show that there exists constants $K > 0$ and $c \in (0, 1)$ such that$|\theta_{n, 1} - \theta_1| \leq Kc^n$?
My attempt: For $n \geq 1$
\begin{align*}
|\theta_{n, 1} - \theta_1| & = \left|\frac{\theta_1(1+\theta_2) - \theta_2\theta_{n-1, 1}}{r_n} - \theta_1\right|\\
& \leq |\theta_1| +  \left|\frac{\theta_1(1+\theta_2) - \theta_2\theta_{n-1, 1}}{r_n}\right|\\
& = |\theta_1| +  \left|\frac{\theta_1 + \theta_2(\theta_1 - \theta_{n-1, 1})}{r_n}\right|\\
& \leq \underbrace{|\theta_1|\left(1 + \frac{1}{r_n}\right)}_{b_n} + \underbrace{\left|\frac{\theta_2}{r_n}\right|}_{d_n} \underbrace{\left|\theta_1 - \theta_{n-1, 1}\right|}_{\delta_{n-1}}\\
\delta_n & \leq b_n + d_n \delta_{n-1}
\end{align*}.
\begin{align*}
\delta_n & \leq b_n + b_{n-1}d_{n} + \dots +b_2d_nd_{n-1}\dots d_3 + d_nd_{n-1}\dots d_{2}\delta_{1}\\
\end{align*}
How can I proceed from here?
 A: For each $n$ put $\Delta_n=\theta_{n,1}-\theta_1$. Then
$$\Delta_n=-\frac{1}{r_n}(\theta_1(r_n-1)+\theta_2\Delta_{n-1}).$$
Now we can show that if $\theta_1>0$ then a sequence $\{\Delta_n \}$ not necessarily converges to zero as quickly as $Kc^n$ for some $c\in (0,1)$. Indeed, consider a sequence $r_n=1+\tfrac1{n+k}$ for some constant $k>0$ such that $r_1<1+\theta_1^2+\theta_2^2$. Then for each $n$
$$\Delta_n=-\frac{1}{1+\tfrac{1}{n+k}}(\theta_1\tfrac 1{n+k}+\theta_2\Delta_{n-1})=$$
$$=-\frac{1}{n+k+1}(\theta_1+(n+k)\theta_2\Delta_{n-1}).$$
For each $n$ put $E_n=(n+k+1)\Delta_n$. If $|\Delta_n|\le Kc^n$ then $\{E_n=(n+k+1)\Delta_n \}$  converges to zero. On the other hand, 
$E_{n}=-(\theta_1+\theta_2E_{n-1}).$
$E_{n+1}=-(\theta_1+\theta_2 E_n).$
$E_{n+1}-E_n=-\theta_2(E_n-E_{n-1})$
$E_{n+1}-E_n=(-\theta_2)^{n-1}(E_2-E_1)$
$E_n=E_1+(E_2-E_1)((-\theta_2)^0+(-\theta_2)^{1}+\dots+(-\theta_2)^{n-2})=$ $E_1+\frac{(E_2-E_1)(1+(-\theta_2)^{n-1})}{1-\theta_2}$
Since $|\theta_2|<1$, $\{E_n \}$ converges to $\tfrac {E_2– \theta_2E_1}{1-\theta_2}$. If $E_2– \theta_2E_1=0$ then, since $E_2=-(\theta_1+\theta_2E_1)$, we have $2\theta_2E_1=\theta_1$. This equality can be avoided by adjusting $k$ or $\theta_{1,1}$. 
