Let $\ x_n$ a sequence definded as follows: $$\begin{align}x_1&=4\\x_{n+1}&= x_n^2 + 5x_n + 9,\quad(n\geq 1)\end{align}$$
Prove that the sequence $\ y_n= \frac{1}{x_1-2}+ \frac{1}{x_2-2}+...+\frac{1}{x_n-2}$ converges and find its limit.
I found that $\ x_n$ is increasing as n increases because $\ x_{n+1}-x_{n}=(x_n-2)^2+5\geq 0,$ but I see no way on how to use this.
Hint
Let $z_n=x_n-3.$ Then we get:
$$\begin{align}z_{n+1}&=(z_n+3)^2-5 (z_n+3)+6\\ &=z_n^2+z_n\end{align}$$
So:
$$\begin{align}\frac {1}{z_{n+1}}&=\frac {1}{z_n (z_n+1)} \\ &=\frac {1}{z_n}-\frac{1} {1+z_n} \\ &=\frac {1}{z_n}-\frac {1}{x_n-2} \end{align}$$
and using telescoping , $$\sum_{i=1}^n\frac {1}{x_n-2}=\frac {1}{z_1}-\frac {1}{z_{n+1}} $$
with $z_1=4-3=1$. in the end, $$x_{n+1}>x_n^2>x_{n-1}^{2^2}>...x_2^{2^{n-1}} $$ thus $$\lim_{\infty}x_n=\lim_{\infty}z_n=+\infty$$
And your sum goes to $1$.