Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n

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Example of Convergent Real Sequence

Let $h, k \in \R_{>0}$.

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

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


Then $\sequence {x_n}$ is convergent to the positive root of the quadratic equation:

$x^2 + x = k$


Proof

First some lemmata:

Lemma 1

$\forall n \in \N_{>1}: k > x_n > 0$

$\Box$

Lemma 2

Consider the subsequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$.

One of them is strictly increasing and the other is strictly decreasing.

$\Box$


From Lemma 2, We have that both $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$ is strictly monotone (one strictly increasing and the other strictly decreasing).

From Lemma 1, they are both bounded above by $k$ and bounded below by $0$.

Hence from the Monotone Convergence Theorem (Real Analysis), they both converge.


Let:

$x_{2 n} \to l$ as $n \to \infty$
$x_{2 n - 1} \to m$ as $n \to \infty$


Then:

\(\displaystyle l\) \(=\) \(\displaystyle \dfrac k {1 + m}\)
\(\displaystyle m\) \(=\) \(\displaystyle \dfrac k {1 + l}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle l + l m\) \(=\) \(\displaystyle k\)
\(\displaystyle m + l m\) \(=\) \(\displaystyle k\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle l\) \(=\) \(\displaystyle m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle l^2 + l\) \(=\) \(\displaystyle k\)

Hence the result.

$\blacksquare$


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