# Convergent Real Sequence/Examples/Term is Geometric Mean of Preceding Two Terms

## Example of Convergent Real Sequence

Let $a, b \in \R_{>0}$ be (strictly) positive real numbers such that $a \le b$.

Let $a, b \in \R_{>0}$ be such that $a \le p \le q \le b$.

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:

$x_n = \begin {cases} p & : n = 1 \\ q & : n = 2 \\ \sqrt{x_{n - 1} x_{n - 2} } & : n > 2 \end {cases}$

That is, beyond the first $2$ terms, each term is the geometric mean of the previous $2$ terms.

Then $\sequence {x_n}$ converges.

## Proof

First note that:

$a \le x_m \le b$ for $m \in \set {1, 2}$.

It can be shown by Proof by Mathematical Induction that:

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

Then:

 $\displaystyle a$ $\le$ $\displaystyle x_{n + 1}$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac a b$ $\le$ $\displaystyle \dfrac {x_{n + 1} } b$ $\displaystyle$ $\le$ $\displaystyle \dfrac {x_{n + 1} } {x_n}$

and:

 $\displaystyle x_{n + 1}$ $\le$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {x_{n + 1} } a$ $\le$ $\displaystyle \dfrac b a$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {x_{n + 1} } {x_n}$ $\le$ $\displaystyle \dfrac b a$

That is:

$\dfrac a b \le \dfrac {x_{n + 1} } {x_n} \le \dfrac b a$

Then we have:

 $\displaystyle {x_{n + 2} }^2 - {x_{n + 1} }^2$ $=$ $\displaystyle x_{n + 1} x_n - {x_{n + 1} }^2$ $\displaystyle$ $=$ $\displaystyle x_{n + 1} \paren {x_n - x_{n + 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x_{n + 2} - x_{n + 1} }$ $=$ $\displaystyle \dfrac {x_{n + 1} } {x_{n + 2} + x_{x + 1} } \size {x_n - x_{n - 1} }$ $\displaystyle$ $\le$ $\displaystyle \dfrac b {a + b} \size {x_n - x_{n - 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x_{n + 2} - x_{n + 1} }$ $=$ $\displaystyle \paren {\dfrac b {a + b} }^{n - 1} \size {x_2 - x_1}$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x_n - x_m}$ $=$ $\displaystyle \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }$ $\displaystyle$ $\le$ $\displaystyle \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \paren {\paren {\dfrac b {a + b} }^{n - 3} + \paren {\dfrac b {a + b} }^{n - 2} + \dotsb + \paren {\dfrac b {a + b} } } \size {x_2 - x_1}$