Question 3.13: Let a > 0. If x1 = 1, and xn+1 = 1/2 (xn + a/xn) for all ...
Let a > 0. If x_{1} = 1, and
x_{n+1} = \frac{1}{2}\left(x_{n} + \frac{a}{x_{n}} \right) for all n ∈ \mathbb{N}, (3.9)
then (x_{n}) converges to \sqrt{a}.
First note that x_{n} > 0 for all n ∈ \mathbb{N}, and that x_{n} satisfies the quadratic equation
t^{2} − 2x_{n+1}t + a = 0,
so its discriminant 4x^{2}_{n+1} − 4a cannot be negative. Therefore x^{2}_{n+1} ≥ a for all n ∈ \mathbb{N}, that is, (x_{n}) is bounded below.
On the other hand, we also have
x_{n+1} − x_{n} = \frac{1}{2}x_{n}+\frac{a}{2x_{n}} − x_{n}
=\frac{a}{2x_{n}} − \frac{x_{n}}{2}
= \frac{a− x^{2}_{n} }{2x_{n}} ≤ 0 for all n ≥ 2.
This implies that (x_{n}) is decreasing, and hence convergent.
Assuming that lim x_{n} = x, we clearly have lim x_{n}+1 = x, and x ≠ 0 because x_{n+1} ≥ \sqrt{a} for all n. Taking the limits of both sides of equation (3.9) now yields
x = \frac{1}{2}\left(x + \frac{a}{x} \right)
2x^{2} = x^{2} + a
x = \sqrt{a},
where the root −\sqrt{a} is excluded, since the sequence is bounded below by 0.