Again! sequences n limits?

Question

Again! sequences n limits?
x_1 and y_1 are given, x_1>y_1>0

x_n+1 = (x_n+y_n)/2
y_n+1 = sqrt(x_n*y_n)

a)proof y_n< y_n+1
b)proof y_1 < x_n+1 < x_n

c)proof 0<(x_n+1) - (y_n+1)< (x_1 - y_1)2^n

d)explain why {x_n} and {y_n} are convergent and have the same limit


anything will help
Thank you all of u guys

Answer 1

If a sequence has a limit, it is convergent.

Answer 2

Claim: x[n] > x[n+1] > y[n+1] > y[n].

This follows from the fact that a > (a + b)/2 > sqrt (a b) > b whenever a > b > 0. Proof of the middle inequality (the rest is trivial):

(a + b)^2 = a^2 + b^2 + 2 a b = (a - b)^2 + 4 a b > 4 a b
Now take square roots and divide by 2 to find
(a + b)/2 > sqrt(4 a b)/2 = sqrt (a b)

Therefore, x[n] is a monotonously decreasing sequence and y[n] is monotonously increasing, and all the time x[n] > y[n]. This proves statements a and b.

That the sequences converge is obvious, because bounded monotonous sequences converge. We must show that they have the same limit, by showing that x[n] - y[n] converges to zero. That this difference is always positive is obvious.

(x[n+1] - y[n+1])^2 = {(x[n] + y[n])/2 - sqrt(x[n] y[n])}^2
... < {(x[n] + y[n])/2}^2 - {sqrt(x[n] y[n])}^2
... = 1/4 x[n]^2 + 1/4 y[n]^2 + 1/2 x[n] y[n] - x[n] y[n]
... = 1/4 x[n]^2 + 1/4 y[n]^2 - 1/2 x[n] y[n]
... = (x[n] - y[n])^2 / 4

So it follows that x[n+1] - y[n+1] < (x[n] - y[n])/2. It is easy to show by induction that
x[n+1] - y[n+1] < (x[1] - y[1]) / 2^n

(I assume that is what you meant... you forgot to divide)

Since the sequence (x[1] - y[1]) / 2^n dominates x[n+1] - y[n+1], and has limit zero, the limit of x[n+1] - y[n+1] must also be zero, therefore x[n] and y[n] converge to the same limit.

Answer 3

It looks like both are just going to infinity. x1 is always bigger than X_n+1, because you are always cutting x1 in half to get x_n+1.

and so on.