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Let there be two positive numbers $a$ and $b$ with $a>b$ and it is given that $a_{0}=a,b_{0}=b$, and we define $$\begin{align} a_{n+1} & =\dfrac{1}{2}(a_{n}+b_{n})\\b_{n+1}& =\sqrt{a_{n}b_{n}}\end{align}$$

Prove that the limit $$\lim_{n\to\infty}2^n\left(a_{n}-b_{n}\right)$$ exists and evaluate it.

This link proves that $\lim\limits_{n\rightarrow\infty}{a_n-b_n}=0$.

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  • $\begingroup$ Its either infinite or zero. $\endgroup$
    – evil999man
    May 21 '14 at 14:45
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A property of the arithmetic-geometric mean is that $\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} b_n$. Also, its convergence rate is superlinear, which means it converges faster than any exponential function. Therefore $\lim_{n \rightarrow \infty} (a_n-b_n)$ goes to $0$ faster than $\lim_{n\rightarrow \infty} 2^n$ goes to $\infty$, and $$ \lim_{n\rightarrow \infty} 2^n(a_n - b_n)=0. $$

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Let $v_n=2^n(a_n-b_n)$. Then $\large \frac{v_{n+1}}{v_n}=\frac{\sqrt{a_n}-\sqrt{b_n}}{\sqrt{a_n}+\sqrt{b_n}}$. As we know that $\sqrt{a_n}+\sqrt{b_n}>b$ which means, $\large 0<\frac{v_{n+1}}{v_n}=\frac{\sqrt{a_n}-\sqrt{b_n}}{\sqrt{a_n}+\sqrt{b_n}}=\frac{{a_n}-{b_n}}{\left(\sqrt{a_n}+\sqrt{b_n}\right)^2}<\frac{{a_n}-{b_n}}{\left(b\right)^2}\longrightarrow 0$. Hence by ratio test we have $v_n\longrightarrow 0$ as $n\longrightarrow \infty$.

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    $\begingroup$ This in fact shows that not only $v_{n} \to 0$, but the series $\sum v_{n}$ is convergent. $\endgroup$ May 22 '14 at 13:38

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