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

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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:

$a \le x_i \le b$ for $i \in \set {1, 2}$.

It can be shown by Proof by Mathematical Induction that:

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

Then:

\(\ds a\) \(\le\) \(\ds x_{n + 1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac a b\) \(\le\) \(\ds \dfrac {x_{n + 1} } b\)
\(\ds \) \(\le\) \(\ds \dfrac {x_{n + 1} } {x_n}\)


and:

\(\ds x_{n + 1}\) \(\le\) \(\ds b\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x_{n + 1} } a\) \(\le\) \(\ds \dfrac b a\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x_{n + 1} } {x_n}\) \(\le\) \(\ds \dfrac b a\)

That is:

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


Then we have:

\(\ds {x_{n + 2} }^2 - {x_{n + 1} }^2\) \(=\) \(\ds x_{n + 1} x_n - {x_{n + 1} }^2\)
\(\ds \) \(=\) \(\ds x_{n + 1} \paren {x_n - x_{n + 1} }\)
\(\ds \leadsto \ \ \) \(\ds \size {x_{n + 2} - x_{n + 1} }\) \(=\) \(\ds \dfrac {x_{n + 1} } {x_{n + 2} + x_{x + 1} } \size {x_n - x_{n - 1} }\)
\(\ds \) \(\le\) \(\ds \dfrac b {a + b} \size {x_n - x_{n - 1} }\)
\(\ds \leadsto \ \ \) \(\ds \size {x_{n + 2} - x_{n + 1} }\) \(=\) \(\ds \paren {\dfrac b {a + b} }^{n - 1} \size {x_2 - x_1}\)


Let $n > m$ be arbitrary. Then:

\(\ds \size {x_n - x_m}\) \(=\) \(\ds \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }\)
\(\ds \) \(\le\) \(\ds \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \paren {\paren {\dfrac b {a + b} }^{n - 3} + \paren {\dfrac b {a + b} }^{n - 4} + \dotsb + \paren {\dfrac b {a + b} }^{m - 2} } \size {x_2 - x_1}\)
\(\ds \) \(<\) \(\ds \paren {\dfrac b {a + b} }^{m - 2} \paren {\sum_{k \mathop = 0}^\infty \paren {\dfrac b {a + b} }^k } \size {x_2 - x_1}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac b {a + b} }^{m - 2} \paren {\dfrac 1 {1 - \frac b {a + b} } } \size {x_2 - x_1}\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \paren {\dfrac b {a + b} }^{m - 2} \paren {\dfrac {a + b} a} \size {x_2 - x_1}\)
\(\ds \) \(\to\) \(\ds 0\) as $m \to \infty$

Therefore $\sequence {x_n}$ is a Cauchy sequence and hence converges.

$\blacksquare$


Sources

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