# Prove that the sequence $\ y_n= \frac{1}{x_1-2}+ \frac{1}{x_2-2}+…+\frac{1}{x_n-2}$ converges and find it`s limit

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: It's pretty easy to show inductively that $x_n\geq 2^n+2.$ – Thomas Andrews Mar 7 '18 at 15:17
• And $x_{n+1}-x_n\neq (x_n-3)^2.$ unless there's a typo in your recurrence relation, and it should be $$x_{n+1}=x_n^2\color{red}{-}5x_n+9.$$ – Thomas Andrews Mar 7 '18 at 15:19
• Something is wrong in $x_{n+1)-x_n$. – hamam_Abdallah Mar 7 '18 at 15:21
• If it is supposed to be $x_{n+1}=x_n^2\color{red}{-}5x_n+9,$ then you can show that $x_{n}\geq 3+n$, and then, using your formula, $x_{n+1}=x_n+(x_n-3)^2$ show that $x_n\geq 3+(n-1)^2$ for all $n.$ – Thomas Andrews Mar 7 '18 at 15:28

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 $1$.

• +1 That assumes the recurrence in the original question was incorrect, but that seems likely. – Thomas Andrews Mar 7 '18 at 15:56