I made a substitution y(n) = 1/x(n)
Then the recurrence becomes:
1/y(n+1) = (1/y(n)^2) / (1/y(n-1) + 2*y(n))
y(n+1) = (1/y(n-1) + 2*y(n)) / (1/y(n)^2)
y(n+1) = y(n)^2/y(n-1) + 2*y(n)
Manually evaluating this for a few terms yields 1,3,15,105,945.... These are the odd double factorials y(n)=(2n-1)!!.
So lets prove it by substituting y(n) = (2n-1)!!:
(2n+3)!! = 2*(2n+1)!! + (2n+1)!!^2/(2n-1)!!
(2n+3)!! = 2*(2n+1)!! + (2n+1)*(2n+1)!!
(2n+3)!! = (2n+3)*(2n+1)!!
(2n+3)!! = (2n+3)!!
Then going back: x(n) = 1/(2n-1)!! for x>=1
Solution
