How can I show following sequence of square root is convergent?

Question

For positive $p$

$$\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}=x$$

I want to show that this is convergent by using contractive mapping.

Can anyone help me?

Answer 1

You have $x=\sqrt{p+\text{something}}$, and "something" is equal to $x$ itself. So

$$ x = \sqrt{p + x}. \tag 1 $$

Thus $x$ is a fixed point of the mapping $$w\mapsto \sqrt{p+w}. \tag 2$$ We can ask whether that mapping is a contraction.

$$ \frac d {dw} \sqrt{p+w} = \frac{ 1 }{2 \sqrt{p+w}}. $$

If the absolute value of the derivative remains less than some number $c<1$, then we can use the mean value theorem to show that this is a contraction. The derivative is less than $c$ iff $2\sqrt{p+w} > 1/c$, and that means $p+w > 1/(4c^2)$. Thus on the interval

$$ \left( \frac 1 {4c^2} - p, \infty \right) \tag 3 $$

the mapping $(2)$ is a contraction, and thus must have exactly one fixed point in that interval.

If we square both sides of $(1)$ we get a quadratic equation $$ x^2 - x - p = 0. $$ The solutions are $$ x = \frac{1 \pm\sqrt{ 1+4p }} 2. $$

Is exactly one of those inside the interval $(3)$?