Convergent Real Sequence/Examples/x n = root x n-1 y n-1, 1 over y n = half (1 over x n + 1 over y n-1)

Example of Convergent Real Sequence
Let $\sequence {x_n}$ and $\sequence {y_n}$ be the real sequences defined as:


 * $x_n = \begin {cases} \dfrac 1 2 & : n = 1 \\ \sqrt {x_{n - 1} y_{n - 1} } & : n > 1 \end {cases}$


 * $\dfrac 1 {y_n} = \begin {cases} 1 & : n = 1 \\ \dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } } & : n > 1 \end {cases}$

Then both $\sequence {x_n}$ and $\sequence {y_n}$ converge to the limit $\dfrac \pi 4$.

Proof
By definition, we have that:
 * $\sqrt {x_{n - 1} y_{n - 1} }$ is the geometric mean of $x_{n - 1}$ and $y_{n - 1}$


 * $\dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } }$ is the reciprocal of the harmonic mean of $x_n$ and $y_{n - 1}$.

We are given that:


 * $\dfrac 1 2 = x_1 < y_1 = 1$

Assuming $x_{n - 1} < y_{n - 1}$, it follows from Geometric Mean of two Positive Real Numbers is Between them that:
 * $x_{n - 1} < x_n < y_{n - 1}$

Given that $x_n < y_{n - 1}$, it follows from Harmonic Mean of two Real Numbers is Between them that:
 * $x_n < y_n < y_{n - 1}$

It follows from the Principle of Mathematical Induction that:


 * $x_{n - 1} < x_n < y_n < y_{n - 1}$ for $n = 2, 3, \ldots$}

Hence we have:
 * $\sequence {x_n}$ is strictly increasing and bounded above by $y_1 = 1$.


 * $\sequence {y_n}$ is strictly decreasing and bounded below by $x_1 = \dfrac 1 2$.

So by the Monotone Convergence Theorem (Real Analysis), both $\sequence {x_n}$ and $\sequence {y_n}$ converge.

Let:
 * $x_n \to l$ as $n \to \infty$
 * $y_n \to m$ as $n \to \infty$

Then:
 * $l^2 = l m$

and:
 * $\dfrac 1 m = \dfrac 1 2 \paren {\dfrac 1 l + \dfrac 1 m}$

from which it follows that:
 * $l = m$

It remains to be shown that $l = m = \dfrac \pi 4$.