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# 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 $x_{n}\geq 1$. By the defining recursions, we have

$x_{{n+1}}={a_{{n+1}}\over g_{{n+1}}}={a_{n}+g_{n}\over 2\sqrt{a_{n}g_{n}}}={1% \over 2}\left(\sqrt{a_{n}\over g_{n}}+\sqrt{g_{n}\over a_{n}}\right)={1\over 2% }\left(\sqrt{x_{n}}+{1\over\sqrt{x_{n}}}\right)$ |

Since $x_{n}\geq 1$, we have $1/\sqrt{x_{n}}\leq 1$, and $\sqrt{x_{n}}\leq x_{n}$, hence

$x_{{n+1}}-1={1\over 2}\left(\sqrt{x_{n}}+{1\over\sqrt{x_{n}}}-2\right)\leq{1% \over 2}({x_{n}}+1-2)\leq{1\over 2}(x_{n}-1).$ |

From this inequality

$0\leq x_{{n+1}}-1\leq{1\over 2}(x_{n}-1),$ |

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

$\lim_{{n\to\infty}}g_{n}=\lim_{{n\to\infty}}a_{n}.$ |

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 $0\leq x_{n}-1\leq(x_{0}-1)/2^{n}$. 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 $y\geq 0$.

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

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

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

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

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

$\displaystyle x$ | $\displaystyle\leq{x^{2}((x-1)^{2}+2)^{2}\over(x+1)^{2}}$ | ||

$\displaystyle\sqrt{x}$ | $\displaystyle\leq{x((x-1)^{2}+2)\over x+1}$ | ||

$\displaystyle{x+1\over x}\sqrt{x}$ | $\displaystyle\leq(x-1)^{2}+2$ | ||

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

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

$x_{{n+1}}-1\leq{1\over 2}(x_{n}-1)^{2}.$ |

From this equation, we may derive the bound

$x_{n}-1\leq\frac{1}{2^{{2^{n}-1}}}(x_{0}-1)^{{2^{n}}}.$ |

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

## Mathematics Subject Classification

33E05*no label found*26E60

*no label found*

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