Thursday

August 21, 2014

August 21, 2014

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

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.

**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 ...

math - I have a question I have been working on since yesterday and I am not ...

algebra 2 - A sequence is defined recursively by a1=1,an=(an-1+1)^2. Write the ...

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 ...

Fibonacci sequence - The Fibonacci sequence a1=1,a2=1,a3=2,a4=3,a5=5,a6=8… is ...

algebra 1 - sqr 10 times sqr 16 over sqr 5

maths - Find the 9term of the sequence 1 ,sqr(2), 2

Algebra 2 Honors - Right now in my class i am learning about sequences and ...

calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...