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Topic: Help needed to show convergence!
Replies: 2   Last Post: May 1, 2014 12:16 PM

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 David C. Ullrich Posts: 3,555 Registered: 12/13/04
Re: Help needed to show convergence!
Posted: May 1, 2014 12:16 PM

On Wed, 30 Apr 2014 23:38:16 -0700 (PDT), Mathman
<brandi.shafer@gmail.com> wrote:

>let x_1 =2 and define x_(n+1) = 1/2 (x_n + 2/x_n) for n >=1. Is x_n convergent?
>If so, find its limit.
>
>By observation I see that x_1=2 x_2=3/2 x_3=17/12 x_4 = 577/408 ... => x_n is monotone decreasing. I also see that x_n is bounded 0 < x_n <=2 for all n. So I know that since it is bounded monotone that it is convergent but I need to be able to show this formally. So I need to show that 0 < s_(n+2) < s_(n+1) <= 2.
>So using induction I'm trying to show that s_(n+2) < s_(n+1) given that s_(n+1) < 2. So I use the formula to compute s_(n+2) = 1/2 ( x_(n+1) + 2 / x_(n+1)) = 1/2((x_(n+1)^2 + 2)/x_(n+1)) = ??? I have also tried plugging in the above formual for x_(n+1) in the formula for x_(n+2) but I just cant seem to get the algebra to get the terms in a way that I can use the induction hypothesis (maybe I have the wrong hypothesis?) to get that x_(n+2) < x_(n+1). Any help with this part would be appreciated. I can easily shown by induction that x_n < 2 for all n. So I can easily show the bounded but I am having trouble formally showing the monotone decreasing part. Any help is appreciated.
>
>I know to find the limit its L = 1/2(L+2/L) => 2L = L + 2/L => 2L^2 = L^2 + 2 => L^2 = 2 => L = +/- sqrt(2). But since all terms are positive we know that L = sqrt(2). So I know that the above is true I just need help formally proving that x_n is monotone decreasing using induction. Maybe induction is not the way to go?

Of course induction is the way to go. The sequence was _defined_
inductively (or more properly, recursively), so induction is your
only chance.

First prove by induction that x_n > 0; that's trivial.

Now _try_ to prove by induction that x_{n+1} < x_n.
Writing x = x_n to make typing easier, you need to
show that

(x + 2/x)/2 < x

(x^2+ 2)/(2x) < x

x^2 + 2 < 2 x^2

(that step depends on first showing x_n > 0)

2 < x^2.

So _first_ show by induction that x_n^2 > 2.
Then that allows you to show by induction
that x_n is decreasing, and you're done.