# Limit of a sequence

• May 24th 2008, 09:41 PM
arbolis
Limit of a sequence
I'm not able to find a limit of a sequence.
It's related to the Babylonian's method to approximate a square root. In our case, we have $\displaystyle a>0$, and we want to approximate $\displaystyle \sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $\displaystyle x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $\displaystyle x_0>0$. I also showed that the sequence $\displaystyle x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\displaystyle \sqrt a$. Now I must prove that the sequence $\displaystyle {x_n}$ converges to $\displaystyle \sqrt a$ when $\displaystyle n$ tends to $\displaystyle +\infty$ and this is precisely what I'm asking you to help me.
• May 24th 2008, 11:05 PM
Moo
Hello,

Quote:

Originally Posted by arbolis
I'm not able to find a limit of a sequence.
It's related to the Babylonian's method to approximate a square root. In our case, we have $\displaystyle a>0$, and we want to approximate $\displaystyle \sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $\displaystyle x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $\displaystyle x_0>0$. I also showed that the sequence $\displaystyle x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\displaystyle \sqrt a$. Now I must prove that the sequence $\displaystyle {x_n}$ converges to $\displaystyle \sqrt a$ when $\displaystyle n$ tends to $\displaystyle +\infty$ and this is precisely what I'm asking you to help me.

:D

$\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

Let L be the limit of the sequence.
$\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

Therefore, from (1), we get :

$\displaystyle L=\frac{L+\frac aL}{2}$

...
Conclude ? (Nod)
• May 25th 2008, 07:46 AM
arbolis
Thanks so much!! I got $\displaystyle L^2=a$, so $\displaystyle L=\sqrt a$ because $\displaystyle a>0$. I feel I'm not used to find limit of sequences of this type. You helped me a lot. (Hi)
I'll try to demonstrate that $\displaystyle x_n$ is bounded below by $\displaystyle \sqrt a$.
• May 25th 2008, 08:58 AM
ThePerfectHacker
Quote:

Originally Posted by Moo
Hello,

:D

$\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

Let L be the limit of the sequence.
$\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

Therefore, from (1), we get :

$\displaystyle L=\frac{L+\frac aL}{2}$

...
Conclude ? (Nod)

Everything else is perfect, you just need to prove that the sequence is convergent.
• May 25th 2008, 08:59 AM
Moo
Quote:

Originally Posted by ThePerfectHacker
Everything else is perfect, you just need to prove that the sequence is convergent.

"Decreasing and bounded below" :)

If I read correctly, he said he had to prove the limit after showing that it was decreasing and bounded.
• May 25th 2008, 09:07 AM
arbolis
I confused Theperfecthacker because I said " I'll try to demonstrate that http://www.mathhelpforum.com/math-he...1720c05d-1.gif is bounded below by http://www.mathhelpforum.com/math-he...3ee61184-1.gif." I assumed the sequence was bounded below, but I had to prove it and I wasn't able. (Now I'll try).