Convergence of sqrt(2+sqrt(sn)) = s_n+1

[zfolwick]

Homework Statement

Show convergence of s_{n+1}= \sqrt{2+\sqrt{s_n}} where s_1 = \sqrt{2}

and that s_n<2 for all n=1,2,3...

Homework Equations

Let {p_n}be a sequence in metrice space X. {p_n} converges to p iff every neighborhood of p contains p_n for all but a finite number of n.

The Attempt at a Solution

I'm only assuming that's the relevant property to know...

s_n+1 >=s_n so increasing.

s_{n+1} > \sqrt{2}

so \frac{1}{s_{n+1}} <\frac{1}{\sqrt{2}}

but 1/s_n+1 is positive so

0< \frac{1}{s_{n+1}} <\frac{1}{\sqrt{2}} so it's bounded.

Since it's bounded and increasing, the sequence is convergent.

 

[SammyS]

Homework Statement

Show convergence of s_{n+1}= \sqrt{2+\sqrt{s_n}} where s_n = \sqrt{2}

and that s_n<2 for all n=1,2,3...

Is this a typo? " where s_n = \sqrt{2} "

Did you mean to write: s_1 = \sqrt{2} instead ?

 

[zfolwick]

yes you are correct. I fixed the typo

 

[Ray Vickson]

Homework Statement

Show convergence of s_{n+1}= \sqrt{2+\sqrt{s_n}} where s_1 = \sqrt{2}

and that s_n<2 for all n=1,2,3...

Homework Equations

Let {p_n}be a sequence in metrice space X. {p_n} converges to p iff every neighborhood of p contains p_n for all but a finite number of n.

The Attempt at a Solution

I'm only assuming that's the relevant property to know...

s_n+1 >=s_n so increasing.

s_{n+1} > \sqrt{2}

so \frac{1}{s_{n+1}} <\frac{1}{\sqrt{2}}

but 1/s_n+1 is positive so

0< \frac{1}{s_{n+1}} <\frac{1}{\sqrt{2}} so it's bounded.

Since it's bounded and increasing, the sequence is convergent.

You could also analyze this as the dynamical system x_{n+1} = f(x_n), where f(x) = sqrt(2 + sqrt(x)), using the technique of "cobweb plots"; see, eg.,
http://www.math.montana.edu/frankw/ccp/modeling/discrete/cobweb/learn.htm or http://en.wikipedia.org/wiki/Cobweb_plot .

RGV