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 Positive 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 $l = m = \dfrac \pi 4$.

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Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.7 \ (4)$