Let $0<a<b$ be real numbers. Define:
$$x_1=a, \;\; x_2=b, \;\; x_{2n+1}=\sqrt{x_{2n}x_{2n-1}}, \;\;x_{2n+2}= \frac{x_{2n}+x_{2n-1}}{2}$$
Prove that the sequence converges and find its limit.
Solution:
We note that $ [x_{2n+1}, x_{2n+2} ] \subset [x_{2n-1}, x_{2n}]$ and
$$x_{2n+2}- x_{2n+1}\leq \frac{x_{2n}-x_{2n-1}}{2}\leq \cdots \leq \frac{x_2-x_1}{2^n} \xrightarrow{n\rightarrow +\infty}0$$
Therefore the sequence converges and its limit is equal to:
$$\lim x_n = \bigcap_{n=1}^{\infty}\left [ x_{2n-1}, x_{2n} \right ]$$
$$x_1=a, \;\; x_2=b, \;\; x_{2n+1}=\sqrt{x_{2n}x_{2n-1}}, \;\;x_{2n+2}= \frac{x_{2n}+x_{2n-1}}{2}$$
Prove that the sequence converges and find its limit.
Solution:
We note that $ [x_{2n+1}, x_{2n+2} ] \subset [x_{2n-1}, x_{2n}]$ and
$$x_{2n+2}- x_{2n+1}\leq \frac{x_{2n}-x_{2n-1}}{2}\leq \cdots \leq \frac{x_2-x_1}{2^n} \xrightarrow{n\rightarrow +\infty}0$$
Therefore the sequence converges and its limit is equal to:
$$\lim x_n = \bigcap_{n=1}^{\infty}\left [ x_{2n-1}, x_{2n} \right ]$$
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