Question

Would you please tell me what's the limit of `(V_n-U_n)` as `n` approaches positive infinity ? PS: `U_(n+1)= sqrt(U_nV_n)` and `V_(n+1)=(U_n+V_n)/2`

Answer 1

`lim_(n->oo)(V_n-U_n) = 0`

Explanation:

Making `U_n=lambda V_n` and `delta_n = V_n-U_n` we have

`delta_(n+1)=V_(n+1)-U_(n+1)=(1-lambda)V_(n+1) = (lambda+1)/2V_n-sqrt(lambda)V_n` or

`V_(n+1)/(V_n)= 1/2(lambda+1-2sqrt(lambda))/(1-lambda)=1/2(1-sqrt(lambda))^2/(1-(sqrt(lambda))^2)=1/2(1-sqrt(lambda))/(1+sqrt(lambda))`

then `delta_(n+1) = V_0(1/2(1-sqrt(lambda))/(1+sqrt(lambda)))^n` so

`lim_(n->oo)delta_n = 0` because `1/2((1-sqrt(lambda))/(1+sqrt(lambda)))<1`

Answer 2

This is my thoughts:

`U_(n+1) =sqrt(U_nV_n)`
`V_(n+1) = 1/2(U_n+V_n)`

Let `delta(n) = V_n - U_n`

Assume the limit exists, and that:
`lim _(n rarr oo) (V_n - U_n) = epsilon`

`:. lim _(n rarr oo) delta(n) = epsilon`

Then it must also be the case that:
`:. lim _(n rarr oo) delta(n+1) = epsilon`

`:. lim _(n rarr oo) (delta(n+1))/(delta(n)) = 1`

Now:
`delta(n+1) = V_(n+1) - U_(n+1)`

`delta(n+1) = (U_n+V_n)/2 - sqrt(U_nV_n)`

`(delta(n+1))/(delta(n)) = { (U_n+V_n)/2 - sqrt(U_nV_n) } / ( V_n - U_n ) = 1`

`(U_n+V_n)/2 - sqrt(U_nV_n) = V_n - U_n`
`U_n+V_n - 2sqrt(U_nV_n) = 2V_n - 2U_n`
`3U_n-V_n = 2sqrt(U_nV_n)`
`(3U_n-V_n)^2 = (2sqrt(U_nV_n))^2`
`9U_n^2 -6U_nV_n + V_n^2 = 4U_nV_n`
`9U_n^2 -10U_nV_n + V_n^2 = 0`
`(9U_n - V_n)( U_n - V_n ) = 0`

`9U_n - V_n = 0 => U_n = 1/9V_n`
`U_n - V_n = 0 => U_n = V_n`

Not sure how this helps, but I welcome comments

Answer 3

`lim n to oo (V_n-U_n)` is non-negative.

Explanation:

`U_(n+1)=sqrt(U_nV_n) to U_(n+1) > 0 to U_n > 0 to V_n > 0`

Use the property of Means, `GM <= AM`.

Here, `sqrt(U_nV_n) <=(U_n+V_n)/2`

It follows that

`U_(n+1)<=V_(n+1)`. So,

`V_n-U_n>=0`, for n=2, 3, 4, ...# So,

`lim n to oo (V_n-U_n)` is non-negative.