Define a sequence X = (xn) recursively as follows.?

Let x1 = 1; x2 = 2; and xn+2 =(xn+1 + xn)/2 for n ∈ N


(a) Use the PMI to prove that xn ∈ [1, 2] for all n ∈ N


(b) Use the PMI to prove that |xn+1 - xn| = 1/2n-1 for all n ∈ N(n+1,n and n-1 are subscripts)


(c) Prove that X is a Cauchy sequence.


(d) Find the limit of X and thoroughly justify your solution. (It is useful and interesting to code this sequence and run a few iterations to see what's happening with the terms.)

Define a sequence X = (xn) recursively as follows.?
The proofs using the PMI are tedious, but straight forward. For example, the inductive step for (a) is: if 1 %26lt;= x_n %26lt;= 2 for all n such that 2 %26lt;= n %26lt; N, then (1 + 1)/2 %26lt;= (x_n + x_(n+1`))/2 = x_(n+2) %26lt;= (2 + 2)/2.





More elegant proofs are available by finding (using standard techniques for solving difference equations) that x_n = (4/3)*(-1/2)^n + 5/3. If n %26gt;= 2, then (-1/2)^n satisfies -1/8 %26lt;= (-1/2) %26lt;= 1/4, so 1 %26lt;= (4/3)*(-1/8) + 5/3 %26lt;= (4/3)*(1/4) + 5/3 = 2. That proves (a).





(b) There appears to be some error here. For example, x_3 = (x_2 + x_1)/2 = (2 + 1)/2 = 3/2, and |x_3 - x_2| = |3/2 - 2| = 1/2 which is not 1/(2*2-1).





(c) Let e %26gt; 0 be given. Choose N such that 1/e %26lt; (3*2^N)/4. Now if n %26gt; m %26gt; N, then |x_n - x_m| = (4/3)*(1/2^m)|(-1)^n/2^(n-m) - (-1)^m| %26lt;= (4/3)*(1/2^m)*2 %26lt; 4/3)*(1/2^N) %26lt; e. Therefore, this is a Cauchy sequence.





(d) Since (-1/2)^n -%26gt; 0, the limit of the sequence is 5/3.