How can I proove that sequence Xn+1=0.5*(a+Xn) ,(0<a<=1) converges and find (limXn) if n goes to infinity ?

Question

How can I proove that sequence Xn+1=0.5(a+Xn) (0

Answer 1

I forgot too much about this stuff to prove convergence. However, if you can take care of that side of things...

Knowing that the sequence converges, then consecutive terms are nearly equal as n grows larger. This means

lim(n to inf) X_(n+1) = X_n

0.5(a+X_n) = X_n

X_n = a.

EDIT: I got something better, you can show that

X_n = a/2 + a/4 + a/8 + ... + a/2^n + X_0/2^n

where X_0 is the initial term. Of course X_0/2^n converges to zero so you're left with a geometric series.

Answer 2

Use the ratio test:

Suppose X0 = 0 and a=1

Then X1 = .5(0+1)= .5
Then X2 = .5(.5+1) = (1/2)*(3/2)=3/4
Then X3 = .5(1+3/4)=(1/2)(7/4= 7/8
Then X4 = .5(1+7/8)= (1/2)(15/8)=15/16
Then X5 = .5(1+15/16) =(1/2)(31/16)= 31/32

Each term remains <1 and so the series converges. The limit as n -> infinity is obviously 1.

Answer 3

Evaluate.

Answer 4

Duh. For converging to x, you'll bound |x_n - x| . For converging to 0, you'll bound |(x_n-x) - 0|. But that's exactly |x_n - x|.