Vrftsjtryk

Let $rho_n$ be a sequence of positive numbers converging to $1$. Fix $lambda > 0$. Let $$u_3 = - rho_2 / lambda$$ and consider the following sequence

$$u_{n+1} = frac{-1}{frac{lambda}{rho_n} - u_n}, quad n geq 4.$$ Notice that $u_3$ is negative and thus $u_n$ is negative for all $n geq 3$. Assuming a limit $u_{infty} = lim_{n to infty} u_n$ exists, one can show (by solving for the negative root of the quadratic equation obtained from $u_{infty}= -1/ (lambda - u_{infty})$) that $u_{infty}= lambda /2 - sqrt{(lambda/2)^2+1}$ which is between $-1$ and $0$ for all positive $lambda$.

Is it true that the limit $u_{infty} = lim_{n to infty} u_n$ exists? If so, how does one prove it?

If one uses the notation for finite truncations of continued fractions, then the above can be represented as $u_{n+1}=-[a_n, a_{n-1},ldots, a_3]$, where the sequence $(a_n)$ converges as $n to infty$. Here $a_n=lambda/rho_n$ for $n geq 4$ and $a_3=lambda/rho_3-u_3$
Here the notation $[a_n, a_{n-1},ldots, a_3]$ means

$$ cfrac{1}{a_n +cfrac{1}{a_{n-1}+cfrac{1}{a_{n-2}+ddots+cfrac{1}{a_3}}}}$$

Does $u_n$ converge for these kinds of "backwards" continued fractions?