Hero's Method/Proof 1

Proof
First we have the following lemma:

Lemma 1
Also:

This is a quadratic equation in $x_n$.

We know that this equation must have a real solution with respect to $x_n$, because $x_n$ has been explicitly constructed by the iterative process.

Thus its discriminant is $b^2 - 4 a c \ge 0$, where:


 * $a = 1$
 * $b = -2 x_{n + 1}$
 * $c = a$

Thus $x_{n + 1}^2 \ge a$.

Since $x_{n + 1}$ it follows that $x_{n + 1} \ge \sqrt a$ for $n \ge 1$.

Thus $x_n \ge \sqrt a$ for $n \ge 2$.

Now, consider $x_n - x_{n + 1}$.

So, providing we ignore the first term (about which we can state nothing), the sequence $\sequence {x_n}$ is decreasing and bounded below by $\sqrt a$.

Thus by the Monotone Convergence Theorem (Real Analysis), $x_n \to l$ as $n \to \infty$, where $l \ge \sqrt a$.

Now we want to find exactly what that value of $l$ actually is.

By Limit of Subsequence equals Limit of Real Sequence we also have $x_{n + 1} \to l$ as $n \to \infty$.

But $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$.

Because $l \ge \sqrt a$ it follows that $l \ne 0$.

So by the Combination Theorem for Sequences:
 * $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2 \to \dfrac {l + \dfrac a l} 2$ as $n \to \infty$

Since a Convergent Real Sequence has Unique Limit, that means:
 * $l = \dfrac {l + \dfrac a l} 2$

and so (after some straightforward algebra):
 * $l^2 = a$

Thus:
 * $l = \pm \sqrt a$

and as $l \ge +\sqrt a$ it follows that:
 * $l = +\sqrt a$

Hence the result.