Fix a number \(\alpha \in \mathbb{R}^+\). Choose \(x_1 > \sqrt{\alpha}\) and define:

\(x_{n+1} = \frac{1}{2}(x_n +\frac{\alpha}{x_n})\)

We will prove that the sequence of all \(x_n\):

(1) Is bounded below

(2)\(\displaystyle \lim_{n \to \infty} x_n = \sqrt{\alpha}\)

(3) Is monotone decreasing

Proof:

(1) We will show that \(x_n > \sqrt{\alpha}\) \(\forall\) \(n \in \mathbb{N}\).

Using induction, we establish a base case as \(n=1\), which is trivial as this is given:

\(x_1 > \sqrt{\alpha}\)

Now, suppose the statement is true for \(n= k-1\):

\(x_{k-1} > \sqrt{\alpha}\)

\(\Rightarrow\) \(x_{k-1} - \sqrt{\alpha} > 0\)

\(\Rightarrow\) \((x_{k-1} - \sqrt{\alpha})^2 > 0\)

\(\Rightarrow\) \(x_{k-1}^2 -2\sqrt{\alpha}x_{k-1} + \alpha > 0\)

\(\Rightarrow\) \(x_{k-1}^2 + \alpha > 2\sqrt{\alpha}x_{k-1}\)

\(\Rightarrow\) \(x_{k-1} +\frac{\alpha}{x_{k-1}} > 2\sqrt{\alpha}\)

\(\Rightarrow\) \(\frac{1}{2} (x_{k-1} + \frac{\alpha}{x_{k-1}}) > \sqrt{\alpha}\)

\(\Rightarrow\) \(x_k > \sqrt{\alpha}\)

Then we may conclude that the statement is true for \(n=1\) and if it is true for some \(n=k-1\) then it is true for \(n=k\). The proof follows by induction.

This proves (1).

(2) Proceeding to take the limit:

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

Now, if this limit exists then it must be true that:

\(\displaystyle \lim_{n \to \infty} x_n = \displaystyle \lim_{n \to \infty} x_{n+1}\)

Let this limit be represented as \(L\). Then we have:

\(L = \frac{1}{2}(L+ \frac{\alpha}{L})\)

\(\Rightarrow\) \(2L^2 = L^2 + \alpha\)

\(\Rightarrow\) \(L = \sqrt{\alpha}\)

Then the sequence is bounded below with \(\displaystyle \lim_{n \to \infty} x_n =\sqrt{\alpha}\)

This proves (2).

(3) Using induction, we establish a base case as \(n=1\):

\(\Rightarrow\) \(x_1 > \frac{1}{2} (x_1 + \frac{\alpha}{x_1})\)

Now we need to actually prove this. Start with:

\(x_1 > \sqrt{\alpha}\)

\(\Rightarrow\) \(\frac{1}{2}x_1^2 > \frac{1}{2} \alpha\)

\(\Rightarrow\) \(x_1^2 - \frac{1}{2}x_1^2 > \frac{1}{2} \alpha\)

\(\Rightarrow\) \(x_1^2 > \frac{1}{2} (x_1^2 +\alpha)\)

\(\Rightarrow\) \(x_1>\frac{1}{2}(x_1 +\frac{\alpha}{x_1})\)

\(\Rightarrow\) \(x_1>x_2\)

The base case clearly holds. Now the inductive case \(n=k\):

In (1) we proved that \(x_n > \sqrt{\alpha}\) \(\forall\) \(n \in \mathbb{N}\). Begin with:

\(x_k > \sqrt{\alpha}\)

\(\Rightarrow\) \(\frac{1}{2} x_k^2 > \frac{\alpha}{2}\)

\(\Rightarrow\) \( x_k^2 > \frac{1}{2}x_k^2 + \frac{\alpha}{2}\)

\(\Rightarrow\) \(x_k > \frac{1}{2}(x_k + \frac{\alpha}{x_k})\)

\(\Rightarrow\) \(x_k > x_{k+1}\)

The proof again follows by induction. This proves (3).

We have shown that the sequence is monotone decreasing and bounded below with a limit at infinity that exists. It follows that the sequence converges.

QED

Note the utility of this sequence. If we let alpha equal any real number, we now have an algorithm that will approximate the square root of alpha, with accuracy increasing as \(n\) increases. The proof above is my own, but the sequence itself is not original. I thought this was interesting.

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## Comments

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TopNewestTo prove (2), you made the assumption that the limit exists. It is not necessarily valid to assume that the limit exists, and then show that the limit is a certain value.

You can actually show that the limit exists using (1) and (3). Note that (3) doesn't use (2). That is often the approach taken for this question.

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I was thinking about this. I read that in order to show that a sequence converges, if it is monotone decreasing, then you must show that it is bounded below, which was done above without (2). So I'm not sure if (2) really needs to be in the proof at all. However, we could have just as easily shown that the sequence is bounded by zero, but the sequence clearly doesn't converge to zero. If (1) and (3) can be used to show that the limit exists, then would (2) be more appropriately placed at the end?

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The standard approach is to

The order of the first 2 steps is interchangeable.

Note that bounded below follows directly from AM-GM.

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