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.
To show that An < 2 for all n, we will use a proof by induction.
First, we will prove the base case: A1 = sqrt(2) < 2. This is clearly true because the square root of 2 is approximately 1.414, which is less than 2.
Now, let's assume that An < 2 for some arbitrary value of n. We will use this assumption to prove that A(n+1) < 2.
By the recursive definition of the sequence, we have A(n+1) = sqrt(2 + An).
Since An < 2 (as assumed), we know that 2 + An < 2 + 2 = 4.
Therefore, sqrt(2 + An) < sqrt(4) = 2.
So we have shown that if An < 2, then A(n+1) < 2.
Since the base case A1 < 2 is true, and the statement holds for A(n+1) whenever it holds for An, we can conclude that An < 2 for all positive integers n.
Therefore, the value of An for any value of n is always less than 2.