# Thread: Limit of a sequence

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

2. Hello,

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

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

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

Therefore, from (1), we get :

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

...
Conclude ?

3. Thanks so much!! I got $L^2=a$, so $L=\sqrt a$ because $a>0$. I feel I'm not used to find limit of sequences of this type. You helped me a lot.
I'll try to demonstrate that $x_n$ is bounded below by $\sqrt a$.

4. Originally Posted by Moo
Hello,

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

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

Therefore, from (1), we get :

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

...
Conclude ?
Everything else is perfect, you just need to prove that the sequence is convergent.

5. 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.

6. I confused Theperfecthacker because I said " I'll try to demonstrate that is bounded below by ." I assumed the sequence was bounded below, but I had to prove it and I wasn't able. (Now I'll try).