Finding a limit of two given series

• Oct 29th 2005, 09:24 AM
hoeltgman
Finding a limit of two given series
I got two series defined as following:

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

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 $\displaystyle \sqrt{ab}$
• Oct 29th 2005, 11:33 AM
hpe
Quote:

Originally Posted by hoeltgman
I got two series defined as following:

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

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 $\displaystyle \sqrt{ab}$

You must assume that $\displaystyle ab > 0$. I guess you have already proved that $\displaystyle \lim_{n \to \infty}u(n) = \lim_{n \to \infty}v(n)$. Call the limit U.
Now from the definition,
$\displaystyle v(n+1) = \frac{2 u(n)v(n)}{u(n)+v(n)} = \frac{u(n)v(n)}{u(n+1)}$.
Therefore for all n,
$\displaystyle u(n+1)v(n+1) = u(n)v(n)$.
By induction,
$\displaystyle u(n)v(n) = u(0)v(0) = ab$
for all n. Now take the limit:
$\displaystyle U^2 = ab$ and therefore $\displaystyle U = \sqrt{ab}$. :D
• Oct 29th 2005, 12:02 PM
hoeltgman
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. ;)