Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$ Given $x_1=2$
and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$
Prove that $y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$ converges and find its limit.


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*To prove a convergence we can just estimate $x_n > n$, therefore $y_n<z_n$, where $z_n=\sum_{k=1}^{n}\frac{1}{k^2-1}$ and $z_n$ converges, then $y_n$ converges too.


*We can notice that $x_n^2+2x_n+5=(x_n+1)^2+4$. So $x_{n+1}$ is one of the roots of the equation:
$x_{n+1}^2-(x_n+1)x_{n+1}-1=0$
So $x_{n+1}^2-1=(x_n+1)x_{n+1}$ and therefore: $y_n=\sum_{k=1}^n \frac{1}{(x_{n-1}+1)x_{n}}$
I'm stuck here.
 A: Using the relation $x_{n+1}^2 - 1 = x_{n+1}(x_n + 1)$, we find that
\begin{align*}
\frac{1}{x_n + 1} - \frac{1}{x_{n+1} + 1}
&= \frac{x_{n+1}}{x_{n+1}^2 - 1} - \frac{1}{x_{n+1} + 1} \\
&= \frac{1}{x_{n+1}^2 - 1}.
\end{align*}
So it follows that
\begin{align*}
y_n
&= \frac{1}{x_1^2 - 1} + \sum_{k=1}^{n-1} \left( \frac{1}{x_k + 1} - \frac{1}{x_{k+1} + 1} \right) \\
&= \frac{1}{x_1^2 - 1} + \frac{1}{x_1 + 1} - \frac{1}{x_n + 1} \\
&\xrightarrow[n\to\infty]{} \frac{x_1}{x_1^2 - 1} = \frac{2}{3}.
\end{align*}
A: Making use of the following equality, and standard partial fraction decompositions: $$x_{k+1}^2-(x_k+1)x_{k+1}-1=0\implies(x_k+1)=\frac{x_{k+1}^2-1}{x_{k+1}}$$
$$\begin{align}y_n&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{x_{k+1}}{x_{k+1}^2-1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{2}\left(\frac{1}{x_{k+1}-1}+\frac{1}{x_{k+1}+1}\right)\right)\\&=\frac{1}{4}+\frac{1}{12}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{3}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{2}y_n\\\implies y_n&=\frac{2}{3}-\frac{1}{2}\frac{1}{x_{n+1}-1}-\frac{1}{2}\frac{1}{x_{n+1}-1}\\\implies y_n&\to\frac{2}{3},\quad n\to\infty\end{align}$$
Convergence is slow, but my Python script indeed evaluates this as $0.66665666$ at $n=100,000$.
