Posted by **Belle** on Sunday, January 7, 2007 at 1:49pm.

Let {An} be the sequence defined recursively by A1=sqr(2) and A(n+1) = sqr(2+An) for n is bigger and equal to 1. Show that An < 2. What is An? and how do I find it? Thank you for your time.

An is the nth number in the sequence, and is defined by the recursion rule

A(n+1) = sqr(2+An)

A1 = sqrt 2 = 1.141..

A2 = sqrt (2 + sqrt 2) = 1.773..

A3 = sqrt [2 + sqrt (2 + sqrt 2)]= 1.994..

A4 = sqrt (2 + 1.994.) = 1.998..

etc.

Your job is to prove that no matter how large n is, An < 2.

An cannot equal or exceed 2 unless A(n-1) equals or exceeds two. You can apply this logic going backwards in n to n=1, and conclude that An never reaches 2.

## Answer This Question

## Related Questions

- Math - Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My...
- Math - Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My ...
- Geometry - Find the area of a hexagon with the indicated apothem: 6 sqr root 3 ...
- math - I have a question I have been working on since yesterday and I am not ...
- Math - this is a difficult question for me please help! thankyou A sequence is ...
- math asap - this is a difficult question for me please help! thankyou A sequence...
- algebra 2 - A sequence is defined recursively by a1=1,an=(an-1+1)^2. Write the ...
- maths - Find the 9term of the sequence 1 ,sqr(2), 2
- Fibonacci sequence - The Fibonacci sequence a1=1,a2=1,a3=2,a4=3,a5=5,a6=8… is ...
- Algebra2/Trig - sqr(x^2-4)+(x^2)/x^2+1 I got ((x^2+1)*sqr(x^2-4)+x^2)/x^2+1 But ...

More Related Questions