Question

已知x0=0,x1=1,xn+1=(xn+xn-1)/2,求n→无穷大时数列xn的极限

Answer 1

x(n+2) = [x(n+1)+x(n)]/2,

x(n+2) - x(n+1) = -[x(n+1)-x(n)]/2,

{x(n+1)-x(n)}是首项为x(1)-x(0)=1,公比为(-1/2)的等比数列.

x(n+1)-x(n) = (-1/2)^n, n=0,1,2,..

x(n+2) + x(n+1)/2 = x(n+1) + x(n)/2,

{x(n+1) + x(n)/2}是首项为x(1)+x(0)/2 = 1,的常数数列.

x(n+1) + x(n)/2 = 1, n=0,1,2,...

1 - (-1/2)^n = [x(n+1)+x(n)/2] - [x(n+1)-x(n)] = 3x(n)/2,

n->无穷大时,3x(n)/2 -> 1 - 0 = 1, x(n) -> 2/3