I have the following sequence
$$x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right),~n>1,~x_1=1$$
How can I prove the sequence is Cauchy, and get the limit?
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Are you sure the sequence is: $ x_n= (\frac{1}{2})(x_n +\frac{x_n}{2})$ and isn't $ x_{n+2}= (\frac{1}{2})(x_{n+1} +\frac{x_n}{2})$? – hamid kamali May 05 '15 at 19:16
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i verified that the sequence is $x_(n+1)=(1/2)(x_n +2/x_n)$ – tania May 05 '15 at 19:55
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That's very different. I edited it for you — is this what you meant? – Akiva Weinberger May 05 '15 at 23:51
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Awesome. Mille merci – tania May 06 '15 at 01:28
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Once you have shown that the sequence is convergent then by the given recursive relation it not hard to find the limit. Just take the limit as $n\longrightarrow \infty$ in both sides and you will get that limit should be $1$. So, it is enough to show that the given sequence is convergent. First note that $x_n\geq 0$ for all $n$. So, if we show that $x_n$ is a decreasing sequence then we will be done, because any decreasing sequence which is bounded below is convergent. Let us now show this \begin{align} x_{n+1} - x_n &= \frac{1}{2x_n}(x_n^2 +2)-x_n\\ &=\frac{2-x_n^2}{2x_n}\\ &=\frac{2-\frac{1}{4x_{n-1}^2}(x_{n-1}^2+2)^2}{\frac{x_{n-1}^2+2}{x_{n-1}}}\\ &= -\frac{x_{n-1}^4+4-x_{n-1}^2}{4x_{n-1}(x_{n-1}^2+2)}\\ &= -\frac{(x_{n-1}^2-2)^2}{4x_{n-1}(x_{n-1}^2+2)}\\ &\leq 0. \end{align} Therefore, $x_{n+1}\leq x_n$ for all $n$.
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