Finding a limit of two given series

**hoeltgman** · October 29th 2005, 09:24 AM

I got two series defined as following:

$u(n+1) = \dfrac{u(n) + v(n)}{2} \text{with u(0)=a}$
$\dfrac{1}{v(n+1)} = \dfrac{1}{2}(\dfrac{1}{u(n)} + \dfrac{1}{v(n)}) \text{with v(0)=b}$

a and b are positive real numbers.

These two series converge to the same limit. Which I'm supposed to find. I managed to prove that these two series are convergent and even adjacent. But I don't see how to get the limit. All I found out by inserting numbers was that the limit probably is $\sqrt{ab}$

**hpe** · October 29th 2005, 11:33 AM

Originally Posted by hoeltgman

I got two series defined as following:

$u(n+1) = \dfrac{u(n) + v(n)}{2} \text{with u(0)=a}$
$\dfrac{1}{v(n+1)} = \dfrac{1}{2}(\dfrac{1}{u(n)} + \dfrac{1}{v(n)}) \text{with v(0)=b}$

a and b are positive real numbers.

These two series converge to the same limit. Which I'm supposed to find. I managed to prove that these two series are convergent and even adjacent. But I don't see how to get the limit. All I found out by inserting numbers was that the limit probably is $\sqrt{ab}$

You must assume that $ab > 0$ . I guess you have already proved that $\lim_{n \to \infty}u(n) = \lim_{n \to \infty}v(n)$ . Call the limit U.
Now from the definition,
$v(n+1) = \frac{2 u(n)v(n)}{u(n)+v(n)} = \frac{u(n)v(n)}{u(n+1)}$ .
Therefore for all n,
$u(n+1)v(n+1) = u(n)v(n)$ .
By induction,
$u(n)v(n) = u(0)v(0) = ab$
for all n. Now take the limit:
$U^2 = ab$ and therefore $U = \sqrt{ab}$ .

**hoeltgman** · October 29th 2005, 12:02 PM

Thx. I had tried finding some results through some sort of induction, but I always ended up with expressions that either far too complicated, or ten kilometers long.

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