# Show that a recursive sequence converges

## Homework Statement

Let ##x,y## be positive numbers. Let ##a_0 = y## and let ##a_n = \frac{(x/a_{n-1})+a_{n-1}}{2}##. Prove that ##(a_n)## is a decreasing sequence with limit ##\sqrt{x}##.

## The Attempt at a Solution

I'm confused about the initial condition being an arbitrary positive real number. This is because it says to prove that ##(a_n)## is a decreasing sequence, but what if ##0<y< \sqrt{x}##? Then ##(a_n)## can't be decreasing...

.Scott
Homework Helper
You are right.
It should be a "converging" sequence.
Or perhaps they should exempt ##a_0## from the sequence.

LCKurtz
Homework Helper
Gold Member
What you can show is ##a_n \ge \sqrt x## for ##n\ge 1## regardless whether or not ##a_0 = y < \sqrt x##, and that ##a_{n+1}\le a_n## for ##n\ge 1##.

What you can show is ##a_n \ge \sqrt x## for ##n\ge 1## regardless whether or not ##a_0 = y < \sqrt x##, and that ##a_{n+1}\le a_n## for ##n\ge 1##.
Here is my argument:

(1) WTS: ##\forall n \ge 1##, ##a_n \ge \sqrt{x}##.

Note that it is true that ##(x - y^2)^2 \ge 0##. This implies that ##x^2-2xy^2+y^4 \ge 0 \implies x^2 + y^4 \ge 2y^2x \implies x+y^2 \ge 2y\sqrt{x} \implies \frac{\frac{x}{y} + y}{2} \ge \sqrt{x} \implies a_1 \ge \sqrt{x}##. So the base case is satisfied. Now, suppose that for some ##k \in \mathbb{N}## we have that ##a_k \ge \sqrt{x}##. Then ##a_{k+1} = \frac{\frac{x}{a_k} + a_k}{2} \ge \frac{\frac{x}{\sqrt{x}} + \sqrt{x}}{2} = \sqrt{x}##. This completes the induction.

(2) WTS: ##\forall n \ge 1##, ##a_n \ge a_{n+1}##. We see that ##a_{n+1} = \frac{\frac{x}{a_n}+a_n}{2} \le \frac{\frac{a^2_n}{a_n}+a_n}{2} = a_n##.

So we see that ##(a_n)## is bounded below and is decreasing, so by the MCT it has a limit, L. Then ##L = \frac{\frac{x}{L}+L}{2} \implies L = \sqrt{x}##.

Stephen Tashi