show that x_n converges to \sqrt{a}

Fix a>0 and let $\displaystyle x_1>\sqrt{a}$. For n >=1, define

$\displaystyle x_{n+1}$ = $\displaystyle 1/2(x_n +a/x_n)$

Show that $\displaystyle (x_n)$ converges and that lim n--> infinity $\displaystyle x_n = \sqrt{a}$.

Here is what I have got so far, $\displaystyle x_{n+1}$= $\displaystyle 1/2(x_n +a/x_n)$ $\displaystyle >= \sqrt{a}$ since $\displaystyle \sqrt{ab}<=(a+b)/2$

Re: show that x_n converges to \sqrt{a}

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**wopashui** Fix a>0 and let $\displaystyle x_1>\sqrt{a}$. For n >=1, define

$\displaystyle x_{n+1}$ = $\displaystyle 1/2(x_n +a/x_n)$

Show that $\displaystyle (x_n)$ converges and that lim n--> infinity $\displaystyle x_n = \sqrt{a}$.

Here is what I have got so far, $\displaystyle x_{n+1}$= $\displaystyle 1/2(x_n +a/x_n)$ $\displaystyle >= \sqrt{a}$ since $\displaystyle \sqrt{ab}<=(a+b)/2$

1. Use induction to show that $\displaystyle x_n < x_{n=1}$.

2. Let the limit be l. Solve $\displaystyle l = \frac{1}{2} \left(l + \frac{a}{l}\right)$.

Re: show that x_n converges to \sqrt{a}

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Originally Posted by

**mr fantastic** 1. Use induction to show that $\displaystyle x_n < x_{n=1}$.

2. Let the limit be l. Solve $\displaystyle l = \frac{1}{2} \left(l + \frac{a}{l}\right)$.

you meant to show $\displaystyle x_n > x_{n+1}$, since the sequence has to be decreasing to get to $\displaystyle \sqrt{a}$

Re: show that x_n converges to \sqrt{a}

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**wopashui** you meant to show $\displaystyle x_n > x_{n+1}$, since the sequence has to be decreasing to get to $\displaystyle \sqrt{a}$

Yes, typo on my part.