x(n+1) = Root(x(n) +c) (x(n+1))^2 = (x(n) +c) Assumption: Let, x(n)>= x(n+1) (x(n+1))^2 -c >= x(n+1) (x(n+1))(x(n+1)-1) >=c so, x(n)(x(n)-1) >=c x(1)(x(1)-1) >=c root(c)*(root(c)-1) >=c c -root(c) >=c This gives, c 0 so contradiction. hence our assumption was wrong so, x(n)= c+root(c) >=c which is contradictory. Hence, assumption was wrong. So, x(n) < 1+root(c) proved. for lim x tends to inf, x(n) tends to x(n+1) as x(n) is continuously increasing with n so, x(n+1) = x(n) = root(c+ x(n)) (x(n))^2 = c + x(n) (x(n))^2 -x(n) -c = 0 x(n) = [1 + root( 1+4c)]/2 as n tends to infinity. Hope this helps :)