Limit of the recursive sequence $a_{n+1} = \frac{1}{2}(a_n + \frac{p}{a_n})$

Question

I'm trying to prove that the sequence with recursive equation

$$a_{n+1} = \frac{1}{2}\left(a_n+\frac{p}{a_n}\right)|a_1>0;p\in\mathbb{N}$$

Converges to $\sqrt{p}$ as $n$ tends to infinity.

At first I thought I could prove that the sequence is monotonic and bounded from above but then I found out by plotting the sequence on my computer that it has a 'bump' right at the beginning.

I also tried to use the confront theorem by using

$$a_{n+1}\leq\frac{1}{2}(a_n+p)$$

which didn't seem to work either.

What should I do?

Answer 1

Let's think about $a_{n+1}-\sqrt p$. It equals $$a_{n+1}-\sqrt p=\frac{a_n^2-2\sqrt pa_n+p}{2a_n} =\frac{(a_n-\sqrt p)^2}{2a_n}\ge0.$$

So from $a_2$ onwards $a_n\ge\sqrt p$. Then its fairly clear that $a_2\ge a_3\ge a_4\ge\cdots \ge\sqrt p$. The $a_n$ tend to a limit. What can it be?

Answer 2

Note that $a_n-a_{n+1}=a_n-\frac{1}{2}(a_n+\frac{p}{a_n})=\frac{a_n^2-p}{a_n}\frac{1}{2} \geq 0$.

So $a_{n+1} \leq a_n, \forall n \geq 2$. So $a_n$ is decreasing and by MCT, lim $a_n$ exist call it as $a$, hence by algebra of limits, $a$ satisfies $a=\frac{1}{2}(a+\frac{p}{a})$ so $a=...?$

Answer 3

Use the Arithmetic Geometric inequality. for $n\geq 1$

$$a_{n+1}=\frac{1}{2}(a_n+\frac{p}{a_n})\geq\sqrt{a_n\cdot\frac{p}{a_n}}=\sqrt{p}\geq0$$

And by this you can prove it's decreasing from $n=2$, as

$$a_{n+1}-a_n=\frac{1}{2}\cdot\frac{a_n^2-p}{a_n}\geq0$$

this the series converges. Finding out the limit is $\sqrt{p}$ is straightfoward, using limit arithmetic and uniqueness.

Answer 4

Hint :

When $n \to \infty$ , $a_{n+1} \approx a_n$

Now you can put $a_{n+1}=a_n$ and solve the quadratic to get $a_n =\sqrt p $ as $n \to \infty$.