Prove that the limit exists and find it

Question

I'm stuck on a question:

Let $\sigma > 0$; $a_1 > 0$; and $a_{n+1} = 1/2*[a_{n}+(σ/a_{n})]$, where $n = 1,2,\ldots$

How do we prove that the limit n → ∞ a{n} exists and how do we find it? I mean, there's no solid number given in the question, so how does one tackle this type of question? I have a hunch on the use of mathematical induction to prove that a{n} is monotonic, followed by the application of the monotone convergence theorem, however I'm stuck because of the variable σ which I don't exactly know how to deal with. Am i going in the right direction? Any help on this would be appreciated

See http://math.stackexchange.com/questions/82682/proof-of-convergence-babylonian-method-x-n1-frac12x-n-fracax-n — karakusc, Sep 23 '14 at 04:30
Use AM-GM inequality show $(a_n)$ is bounded and apply monotone convergent theorem. — Bumblebee, Sep 23 '14 at 05:49

score 1 · Answer 1 · answered Sep 23 '14 at 06:16

We can use that AM-GM inequality to prove that for every $n \in \mathbb{N}$:

$$ a_{n+1} = \frac{1}{2}(a_{n}+\frac{\sigma}{a_{n}}) \geq \sqrt\sigma $$

So the sequence is limited below. Notice that, if $a_{1}=\sqrt\sigma$, then the whole sequence would be constant and equal to $\sqrt\sigma$, which means that it would be convergent to $\sqrt\sigma$. Assume, then that $a_{1}>\sqrt\sigma$. We first note that $a_{n}>\sqrt\sigma$ for $n\geq2$ (you can prove this using induction, it's very simple).

We affirm that this sequence is monotonically decreasing. We have, using the fact that $a_{n}>\sqrt\sigma$,

$$ a_{n+1}=\frac{a_{n}}{2}+\frac{\sigma}{2a_{n}} <\frac{a_{n}}{2} + \frac{a^{2}_{n}}{2a_{n}} = a_{n} $$

So, the sequence is monotonically decreasing and it's limited below, which means that it is convergent. Since now you know the limit exists, you can just apply the limit on both sides of the equation that defines the sequence to find it (let $\lim a_{n+1}=\lim a_{n}=a$):

$$ \lim a_{n+1} = \frac{1}{2}(\lim a_{n}+\frac{\sigma}{\lim a_{n}}) \to a = \frac{1}{2}(a+\frac{\sigma}{a}) \to a = \lim a_{n} = \sqrt\sigma $$

So, the limit is $\sqrt\sigma$ for every possible value of $a_{1}$. The operations used (sum of limits = limit of sums, constants out and division) to find the limit are all justified because we know that the limit exists and is not zero (all the terms of the sequence are bounded below by a positive constant).

Sorry for being too long in steps that are straightforward, for the most part. Hope this helps!

Prove that the limit exists and find it

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