# 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.

Last edited:

## 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 ?

yes you are correct. I fixed the typo

## 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 [Broken] or http://en.wikipedia.org/wiki/Cobweb_plot .

RGV

Last edited by a moderator: