How is the sequence $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ going to converge to 1 Having some trouble understanding that if $x_{1} = 4$ and the sequence where n is defined as $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ how is it going to converge to 1. 
I have solved using the L as limit and using the quadratic i get two possibilites that are 5, or 1. 
 A: By induction, $1< x_n\le 4$ for all $n$. Indeed, this is true for $n=1$, and if $1< x_n\le 4$, then $1< x_n^2\le 16$ and so $x_{n+1}=\frac{x_n^2+5}{6}$ is $> \frac{1+5}6=1$ and $\le \frac{16+5}6<4$.
With these bounds in mind, we have
$$x_{n+1}-x_n=\frac{x_n^2-6x_n+5}6=\frac{(x_n-1)(x_n-5)}6<0$$
because the first factor in the numerator is positive and the second is negative.
So $\{x_n\}_n$ is bounded and strictly decreasing, hence convergent. You already found out that the only possible limits are $1$ and $5$. However, $5$ can be excluded according to the above observations.
A: You may reason as follows:


*

*Set $f(x) = \frac{x^2+5}{6}$.

*$\Rightarrow f(1) = 1, f(5) = 5$ and for any $a \in (1,5)$ you have $f(a) \in (1,5)$.

*$f''(x) = \frac{1}{3} > 0$, so $f$ is strictly convex.

*$\Rightarrow$ the secant through $(1,1)$ and $(5,5)$ lies above the graph of $f$:
$$f(a) < a \mbox{ for any } a \in (1,5) \Rightarrow (x_n)_{n \in \mathbb{N}} \mbox{ is decreasing.}$$
It follows from your calculation that $L=1$ must be the limit (and cannot be $5$).

A: Since
$$
x_{n+1}=\frac{x_n^2+5}6\tag1
$$
we have
$$
x_{n+2}-x_{n+1}=\frac{x_{n+1}^2-x_n^2}6=(x_{n+1}-x_n)\,\frac{x_{n+1}+x_n}6\tag2
$$
This is a contraction when $\left|\frac{x_{n+1}+x_n}6\right|\lt1$. That is.
$$
-1\lt\frac{(x_n+3)^2-4}{36}\lt1\tag3
$$
Inequality $(3)$ is satisfied when
$$
-3-2\sqrt{10}\lt x_n\lt-3+2\sqrt{10}\tag4
$$
$x_1=4$ is outside the range specified in $(4)$.
