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

Question

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.