Prove convergence of hyperbolic recursive series How to prove that the series $\{x_n\}$ converges, and find its limit, given:
$$ \frac{1}{x_n^2} = \frac{1}{a^2} + \frac{1}{b^2+x^2_{n-1}}$$
I think, to prove convergence for recursive series, we can prove that it has a bound (I think this one has upper bound), and monotonicity. For monotinicity, we should be able to do it by deduction (assuming it holds for $n$). I am having a hard time transforming the series into a form that can be analyzed. If $b = 0$, I can prove it, but in this case, $b$ and $a$ are real positives.
EDIT: (adding my own thoughts and approach)
Consider a hyperbolic curve $f(x)=\frac{1}{x^2}$, then we could represent this recursive series with hyperbolic trigonometric functions. That is, assume $\alpha$ and $\alpha'$ are twice the angle of the vector $\{x_{n-1}^2, \frac{1}{x_{n-1}^2}\}$ and $\{x_{n}^2, \frac{1}{x_{n}^2}\}$ respectively. See image below.

Now, with this reparametrization, we could transform the original series for $x_{n-1} \rightarrow x_{n}$ into a series for $\alpha \rightarrow \alpha'$. With some simple trigonometry, we have:
$$ \left(cosh(\alpha) + b^2\right)^2 - \left(cosh(\alpha') + \frac{1}{a^2}\right)^2 = 1. $$
I think, this could again be reparameterized with trigonometry to prove that $\alpha \rightarrow \alpha'$ converges. However, I think I am stuck here. Maybe we should expand the this expression into series, or maybe use imaginary representation. Any idea for the proof?
 A: Let us examine the series
$$ y_{n+1} = \frac{ay_n + b}{cy_n + d}$$
We can split this into two series (numerator and denominator)
as
$$y_n = \frac{p_n}{q_n}$$
and
$$p_{n+1} = ap_n + bq_n$$
$$q_{n+1} = cp_n + dq_n$$
This can be rewritten in matrix form as
$$\begin{bmatrix}a&b\\c&d \end{bmatrix}\begin{bmatrix}p_n\\q_n\end{bmatrix} = \begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix}$$
And thus
$$ A^n \begin{bmatrix}p_0\\q_0\end{bmatrix} = \begin{bmatrix}p_n\\q_n\end{bmatrix}$$
where
$$A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$$
If $A$ is diagonalizeable then we get that both $p_n$ and $q_n$ are of the form
$$\alpha \lambda_1^n + \beta \lambda_2^n$$
where $\lambda_i$ are the eignevalues (and $\alpha$ for $p_n$ is different from $\alpha$ for $q_n$).
There might be corner cases when the eigenvalues are equal, in which case I believe the general form is $$\alpha n \lambda^n$$ (not sure, though).
In your case the matrix $A$ is
$$A = \begin{bmatrix}1&b^2\\1&a^2+b^2 \end{bmatrix}$$
Which is diagonalizable, and the corner case of equal eignevalues occurs when $(a^2 + b^2)^2 = 4a^2$
In either case, if $\lambda_1$ is the greater root, then the sequence converges to the ratio of the $\alpha$(i.e. coefficients of $\lambda_1$) for $p_n$ and $q_n$.
