# Monotone sequences and Cauchy sequences

• Mar 19th 2009, 11:35 AM
noles2188
Monotone sequences and Cauchy sequences
Suppose that x>0. Define a sequence (s_n) by s_1=k, and s_(n+1) = (((s_n)^2)+x)/(2s_n) for n elements in N. Prove that for any k>0, lim s_n = sqrt(x).

I've gotten that I need to show that (s_(n+1))^2)-x = (((s_n)^2)-x)^2/(4(s_n)^2)=>0, so that s_n=>sqrt(x) for n=>2.

Also I need to prove that (s_n) is decreasing for n=>2, by showing that (s_n)-(s_(n+1))>=0.

Unfortunately I don't know how to exactly show these two steps.

Any help would be great.
• Mar 20th 2009, 08:55 AM
ThePerfectHacker
We will assume $n\geq 2$ and so $s_n = \tfrac{1}{2}(s_{n-1} + xs_{n-1}^{-1})$. Therefore, $s_n^2 = \tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$. We get from here $s_n^2 = x + \tfrac{1}{4}(s_{n-1} - xs_{n-1}^{-1})^2 \geq x \implies s_n \geq \sqrt{x}$. This was your first half of your problem.

Now, $s_{n+1} - s_n = \tfrac{1}{2} (s_n + xs_{n}^{-1}) - s_n = \tfrac{1}{2}(xs_n^{-1} - s_n) < \tfrac{1}{2}(s_n - s_n) = 0$.
(The last inequality follows because $s_n^2 > x \implies s_n > xs_n^{-1}$)
• Mar 21st 2009, 05:05 PM
xalk
Quote:

Originally Posted by ThePerfectHacker
We will assume $n\geq 2$ and so $s_n = \tfrac{1}{2}(s_{n-1} + xs_{n-1}^{-1})$. Therefore, $s_n^2 = \tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$. We get from here $s_n^2 = x + (s_{n-1} - xs_{n-1})^2 \geq x \implies s_n \geq \sqrt{x}$. This was your first half of your problem.

Now, $s_{n+1} - s_n = \tfrac{1}{2} (s_n + xs_{n}^{-1}) - s_n = \tfrac{1}{2}(xs_n^{-1} - s_n) < \tfrac{1}{2}(s_n - s_n) = 0$.
(The last inequality follows because $s_n^2 > x \implies s_n > xs_n^{-1}$)

We have :

$s_n^2 = \tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$.

AND

$s_n^2 = x + (s_{n-1} - xs_{n-1})^2$.

But . $\tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$,is not equal to
$x + (s_{n-1} - xs_{n-1})^2$ as your calculations show .

Please explain your calculations ,i am completely confused
• Mar 21st 2009, 08:59 PM
ThePerfectHacker
Quote:

Originally Posted by xalk
We have :

$s_n^2 = \tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$.

AND

$s_n^2 = x + (s_{n-1} - xs_{n-1})^2$.

But . $\tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$,is not equal to
$x + (s_{n-1} - xs_{n-1})^2$ as your calculations show .

Please explain your calculations ,i am completely confused

I fixed my post now, there was a mistake in it.