1

$X_0 := 2$

and for $X_n$:

$X_{n+1} = \frac12 X_n + \frac1{X_n}$

I know that the sequence is monotone decreasing and I am not sure how to find its limit maybe with Cauchy and partial sums but still I cannot find the exact value of $\lim X_n$ for $n \rightarrow \infty$

Martin Sleziak
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grapher
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  • Almost the same as http://math.stackexchange.com/questions/930484/how-to-prove-this-sequence-is-a-cauchy-sequence –  Dec 18 '14 at 06:59

2 Answers2

6

Hint:

$x_{n+1}$ is a subsequence of $x_n$, so they have the same limit as $n \rightarrow \infty$, provided the limit exists. You already know the sequence is monotone decreasing, so now you'll want to show that it is also bounded below. Once you have that, then a limit $L \neq \pm \infty$ exists by the monotone convergence theorem. At this point, you can simply take a limit of both sides and apply limit arithmetic:

$$\lim(x_{n+1}) = \lim \left( \frac{x_n}{2} + \frac{1}{x_n} \right) = \frac{\lim(x_n)}{2} + \frac{1}{\lim(x_n)}$$

In other words:

$$L = \frac{L}{2} + \frac{1}{L}$$

And now it's just a matter of solving for $L$.

Kaj Hansen
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0

$$\begin{array}{l} \mathop {\lim }\limits_{n \to + \infty } x_{n + 1} = \mathop {\lim }\limits_{n \to + \infty } x_n = \ell \\ \Leftrightarrow \ell = \frac{\ell }{2} + \frac{1}{\ell } \\ \Leftrightarrow \ell = \frac{{\ell ^2 }}{{2\ell }} + \frac{2}{{2\ell }} \\ \Leftrightarrow 2\ell ^2 - \ell ^2 - 2 = 0 \\ \Leftrightarrow \ell \in \left\{ {\frac{{1 - \sqrt {17} }}{4};\frac{{1 + \sqrt {17} }}{4}} \right\} \\ \end{array}$$

but the positive result that is to say, $${\ell = \frac{{1 + \sqrt {17} }}{4}}$$ finally$$ {\mathop {\lim }\limits_{n \to + \infty } x_n = \frac{{1 + \sqrt {17} }}{4}} $$

sabachir
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