TITLE: Evaluating $\lim_{n\to\infty}2^n\left(a_{n}-b_{n}\right)$ for $ a_{n+1} =\dfrac{1}{2}(a_{n}+b_{n})$ and $b_{n+1} =\sqrt{a_{n}b_{n}}$
QUESTION [5 upvotes]: 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 $\lim\limits_{n\rightarrow\infty}{a_n-b_n}=0$.

REPLY [3 votes]: 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 $n\longrightarrow \infty$.