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Let $\sigma > 0$, $a_1>0$ and $a_{n+1}=\frac12(a_n+\frac{\sigma}{a_n})$ where n = 1, 2...

Prove that $\lim_{x\to \infty} a_n$ exists.

The idea that I had was to show that it is an increasing / decreasing sequence and bounded. Then I will use the Monotone Convergent Theorem to show that the sequence converges and hence the limit exists.

However as I did my first calculation of $a_2 - a_1$, I get $a_2 - a_1 = \frac{\sigma - a_1^2}{2a_1}$ . Does this mean that the I have to consider the various cases ($\sigma = a_1^2$, $\sigma>a_1^2$, $\sigma<a_1^2$)? Or am I approaching the question from the wrong direction?

Thanks in advance for the help!

Jason
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    See http://math.stackexchange.com/questions/82682/proof-of-convergence-babylonian-method-x-n1-frac12x-n-fracax-n – karakusc Sep 25 '14 at 04:52

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May be, you could be interested by the fact that $$a_{n+1}=\frac12(a_n+\frac{\sigma}{a_n})$$ looks very much as a Newton iterate of the equation $$f(x)=x^k-y=0$$ Starting from a guess $x_1$ and applying $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ you then have $$x_{n+1}=x_n-\frac{x_n^k-y}{kx_n^{k-1}}=\frac{k-1}{k}x_n+\frac{y}{k}\frac{1}{x_n^{k-1}}$$ So, if you replace $k=2$ and $y=\sigma$, you look for the solution of $x^2=a$ .

Also interesting is that the general solution of the recurrence relation has a solution which write $$a_n=\sqrt{\sigma } \coth \left(2^{n-1} \coth ^{-1}\left(\frac{{a_1}}{\sqrt{\sigma }}\right)\right)$$ which, for infinite value of $n$, has a limit equal to $$\lim_{n\to \infty} a_n=\sqrt{\sigma }$$

Claude Leibovici
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karakusc provided a link to a solution. But the answers there are using different (and more convoluted) techniques from what you proposed. I'm adding an answer here to continue with your good idea to try to show that $(a_n)$ is monotone and bounded.

Note that $$\frac{a_{n+1}}{a_n} = \frac{1}{2}\left(1 + \frac{\sigma}{a_n^2}\right)$$ which shows that $a_{n+1} \leq a_n$ if and only if $a_n^2 \geq \sigma$.

Now, we can rearrange the recurrence as $$2a_{n+1}a_n = a_n^2 + \sigma$$ Then note that $$\begin{align} (a_{n+1} - a_n)^2 &= a_{n+1}^2 - 2a_{n+1}a_n + a_n^2 \\ &= a_{n+1}^2 - a_n^2 - \sigma + a_n^2 \\ &= a_{n+1}^2 - \sigma \\ \end{align}$$ Since the left hand side is nonnegative, so is the right hand side, so this shows that $a_{n+1}^2 \geq \sigma$ for all $n \geq 1$, i.e., $a_n^2 \geq \sigma$ for all $n \geq 2$.

Combining this with the observation in the first paragraph, we see that $a_{n+1} \leq a_n$ for all $n \geq 2$.

We can also show that $(a_n)$ is bounded below as follows. We are given that $a_1 > 0$. The recurrence shows that if $a_n > 0$ then $a_{n+1} > 0$. Therefore by a simple induction, $a_n > 0$ for all $n$. So $0$ is a lower bound. But we can do better than that: we also are given that $\sigma > 0$. Therefore, $a_n^2 \geq \sigma$ for $n \geq 2$ implies $a_n \geq \sqrt{\sigma}$ for $n \geq 2$.

Therefore, for $n \geq 2$, we have shown that $(a_n)$ is decreasing and bounded below (by $\sqrt{\sigma}$), hence convergent.

The limit $L$ therefore exists, and the recurrence shows that it must satisfy $$L = \frac{1}{2}\left(L + \frac{\sigma}{L}\right)$$ so $L = \sqrt{\sigma}$.