Suppose that when n=k, xk
Then when n=k+ 1, x (k+ 1) = √ (2+xk) < √(2+2)=2 holds.
So for any n, xn
Because x(n+ 1)=√(2+xn)>0, so 0.
√(2/2^2+ 1/2)= 1
So x (n+ 1) >: Xn, that is, {xn} monotonically increases.
To sum up, {xn} is monotonically bounded, that is, {xn} limit exists.
Let the limit of {xn} be a, and then find the limit on both sides of x (n+ 1)=√(2+xn).
A=√(2+A)
A^2-A-2=0
(A-2)(A+ 1)=0
A=2 or-1 (truncated)
So the limit of {xn} is 2.