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- #1

- Apr 13, 2013

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I have a question.

It is given that $a>0 , x_{1}=x>0 \text{ and } x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_{n}})$ and I have to show that the sequence $(x_{n})$,at least from its second term,is decreasing and bounded from below from $\sqrt{a}$.Also,I have to find the limit $\lim_{n \to \infty}x_{n}$.

To show that the sequence is decreasing,I thought that I could take $\frac{x_{n+1}}{x_{n}}$ and show that it is smaller that $1$.I found that $\frac{x_{n+1}}{x_{n}}=\frac{1}{2}(1+\frac{a}{x_{n}^{2}})=\frac{1}{2}+\frac{a}{2x_{n}^{2}}$.

But do we know that it is $\frac{1}{2}+\frac{a}{2x_{n}^{2}}<1 \Rightarrow x_{n}>\sqrt{a}$ ?