# Sequence based on arithmetic and geometric means problem

• Oct 29th 2012, 07:01 PM
lm1988
Sequence based on arithmetic and geometric means problem
Let 0 < y1 < x1 and set
xn+1 = (1/2)(xn + yn) and yn+1 = sqrt(xnyn)

Prove that 0 < xn+1 - yn+1 < (x1 - y1)/2n for n in N

Provided solution:

xn+1 - yn+1 = (1/2)(xn + yn) - sqrt(xnyn) < (1/2)(xn + yn) - y = (1/2)(xn - yn)

Hence by induction and by the fact that 0 < yn < xn for n in N, 0 < xn+1 - yn+1 < (x1 - y1)/2n

What I do not understand is the last part of this explanation. I understand that, by induction, xn+1 - yn+1 < (x1 - y1)/2, but why 2n?

• Oct 29th 2012, 08:57 PM
Salahuddin559
Re: Sequence based on arithmetic and geometric means problem
Simple, \[x_{n+1} - y_{n+1} = (1/2)(x_{n} - y_{n})\], and since \[(x_{n} - y_{n}) = 1/2 * (x_{n-1} - y_{n-1})\], \[x_{n+1} - y_{n+1} = (1/2)(x_{n} - y_{n}) = 1/2* 1/2 * (x_{n-1} - y_{n-1}) = 1/4 * (x_{n-1} - y_{n-1} \], and so on. Got the idea?

Salahuddin
Maths online
• Oct 29th 2012, 08:59 PM
chiro
Re: Sequence based on arithmetic and geometric means problem
Hey lm1988.

You have x_(n+1) - y_(n+1) < 1/2(x_n - y_n), but this means x_(n) - y(n) < 1/2(x_(n-1) - y_(n-1)) which means x_(n+1) - y_(n+1) < 1/2^2(x_(n-1) - y(n-1)) and if you repeat this until you get x1 - y1 you get that result.

So basically you have to chain these together.