# convergence of arithmetic-geometric mean

In this entry, we show that the arithmetic-geometric mean^{} converges.
By the arithmetic-geometric means inequality, we know that the sequences^{}
of arithmetic^{} and geometric means^{} are both monotonic and bounded^{}, so
they converge individually. What still needs to be shown is that they
converge to the same limit.

Define ${x}_{n}={a}_{n}/{g}_{n}$. By the arithmetic-geometric inequality^{}, we
have ${x}_{n}\ge 1$. By the defining recursions, we have

$${x}_{n+1}=\frac{{a}_{n+1}}{{g}_{n+1}}=\frac{{a}_{n}+{g}_{n}}{2\sqrt{{a}_{n}{g}_{n}}}=\frac{1}{2}\left(\sqrt{\frac{{a}_{n}}{{g}_{n}}}+\sqrt{\frac{{g}_{n}}{{a}_{n}}}\right)=\frac{1}{2}\left(\sqrt{{x}_{n}}+\frac{1}{\sqrt{{x}_{n}}}\right)$$ |

Since ${x}_{n}\ge 1$, we have $1/\sqrt{{x}_{n}}\le 1$, and $\sqrt{{x}_{n}}\le {x}_{n}$, hence

$${x}_{n+1}-1=\frac{1}{2}\left(\sqrt{{x}_{n}}+\frac{1}{\sqrt{{x}_{n}}}-2\right)\le \frac{1}{2}({x}_{n}+1-2)\le \frac{1}{2}({x}_{n}-1).$$ |

From this inequality

$$0\le {x}_{n+1}-1\le \frac{1}{2}({x}_{n}-1),$$ |

we may conclude that ${x}_{n}\to 1$ as $n\to \mathrm{\infty}$, which , by the definition of ${x}_{n}$,
is equivalent^{} to

$$\underset{n\to \mathrm{\infty}}{lim}{g}_{n}=\underset{n\to \mathrm{\infty}}{lim}{a}_{n}.$$ |

Not only have we proven that the arithmetic-geometric mean converges, but we can infer a rate of convergence from our proof. Namely, we have that $0\le {x}_{n}-1\le ({x}_{0}-1)/{2}^{n}$. Hence, we see that the rate of convergence of ${a}_{n}$ and ${g}_{n}$ to the answer goes as $O({2}^{-n})$.

By more carefully bounding the recursion for ${x}_{n}$ above, we may obtain better estimates of the rate of convergence. We will now derive an inequality. Suppose that $y\ge 0$.

$0$ | $\le {y}^{5}+{y}^{4}+4{y}^{3}+3{y}^{2}$ | ||

${y}^{2}+4y+4$ | $\le {y}^{5}+{y}^{4}+4{y}^{3}+4{y}^{2}+4y+4$ | ||

${(y+2)}^{2}$ | $\le (y+1){({y}^{2}+2)}^{2}$ |

Set $x=y+1$ (so we have $x\ge 1$).

${(x+1)}^{2}$ | $\le x{({(x-1)}^{2}+2)}^{2}$ | ||

$x$ | $\le {\displaystyle \frac{{x}^{2}{({(x-1)}^{2}+2)}^{2}}{{(x+1)}^{2}}}$ | ||

$\sqrt{x}$ | $\le {\displaystyle \frac{x({(x-1)}^{2}+2)}{x+1}}$ | ||

$\frac{x+1}{x}}\sqrt{x$ | $\le {(x-1)}^{2}+2$ | ||

$\frac{1}{2}}\left(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}}\right)$ | $\le 1+{\displaystyle \frac{1}{2}}{(x-1)}^{2}$ |

Thus, because ${x}_{n+1}=(\sqrt{{x}_{n}}+1/\sqrt{{x}_{n}})/2$, we have

$${x}_{n+1}-1\le \frac{1}{2}{({x}_{n}-1)}^{2}.$$ |

From this equation, we may derive the bound

$${x}_{n}-1\le \frac{1}{{2}^{{2}^{n}-1}}{({x}_{0}-1)}^{{2}^{n}}.$$ |

This is a much better bound! It approaches zero far more rapidly
than any exponential function^{}, so we have superlinear convergence.

Title | convergence of arithmetic-geometric mean |
---|---|

Canonical name | ConvergenceOfArithmeticgeometricMean |

Date of creation | 2013-03-22 17:09:46 |

Last modified on | 2013-03-22 17:09:46 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 13 |

Author | rspuzio (6075) |

Entry type | Theorem^{} |

Classification | msc 33E05 |

Classification | msc 26E60 |