# Convergent Real Sequence/Examples/x (n+1) = x n^2 + k

## Example of Convergent Real Sequence

Let $\sequence {x_n}$ be the real sequence defined as:

$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$

where:

$0 < k < \dfrac 1 4$
$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation: $x^2 - x + k = 0$.

Then $\sequence {x_n}$ is convergent such that:

$\displaystyle \lim_{n \mathop \to \infty} x_n = a$

## Proof

First some lemmata:

### Lemma 1

$\forall n \in \N_{>0}: a < x_n < b$

$\Box$

### Lemma 2

Let $a$ and $b$ be the roots of the quadratic equation:

$(1): \quad x^2 - x + k = 0$

Let:

$0 < k < \dfrac 1 4$

Then $a$ and $b$ are both strictly positive real numbers.

$\Box$

From Lemma 1 and Lemma 2 we have that:

$0 < a < x_n < b$

for all $n \in \N_{>0}$.

Then:

 $\displaystyle x_{n + 1} - x_n$ $=$ $\displaystyle {x_n}^2 - x_n + k$ $\displaystyle$ $<$ $\displaystyle 0$ Sign of Quadratic Function Between Roots $\displaystyle \leadsto \ \$ $\displaystyle x_{n + 1}$ $<$ $\displaystyle x_n$

Hence:

$0 < a < x_{n + 1} < x_n < b$

Thus $\sequence {x_n}$ is decreasing and bounded below by $a$.

Hence by the Monotone Convergence Theorem (Real Analysis), $\sequence {x_n}$ converges to its infimum.

It remains to be shown that $\map \inf {x_n} = a$.

Suppose that:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$

Thus:

 $\displaystyle \lim_{n \mathop \to \infty} x_n$ $=$ $\displaystyle l$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} x_{n + 1}$ $=$ $\displaystyle l$ $\displaystyle \leadsto \ \$ $\displaystyle l^2 + k$ $=$ $\displaystyle l$ as $x_{n + 1} = {x_n}^2 + k$ $\displaystyle \leadsto \ \$ $\displaystyle l^2 - l + k$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle l$ $=$ $\displaystyle a$ $\, \displaystyle \text {or} \,$ $\displaystyle l$ $=$ $\displaystyle b$

But $b$ cannot be the infimum of $\sequence {x_n}$ because it is not a lower bound.

Hence:

$\displaystyle \lim_{n \mathop \to \infty} x_n = a$

and the result follows.

$\blacksquare$