Prove the convergence of $x_{n+1}=\sqrt{\alpha +x_n}$, with $x_0=\sqrt \alpha$ using fixed point theorem Consider the well known sequence $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0$.
How would one use the fixed point theorem to prove that it converges to $\sqrt{a} \ \ \ \forall a>0?$
Trying to see whether $f(x) = \sqrt{\alpha+x}$ is a contructor yields following:
$$|f(x)-f(y)|=|\frac{1}{2\sqrt{\alpha+\xi}}| |x-y| \le \frac{1}{2\sqrt{\alpha}} |x-y|,$$
which suggests that $f(x)$ is a contractor only when $\sqrt{\alpha} > \frac{1}{2} $ or $\alpha>\frac{1}{4}$.  But this is obviously false. What's wrong here?
 A: For the fixed point theorem you need a function f that goes from a closed and bounded space in itself, i. e., $ f: [a, b] \longrightarrow [a, b] $. Usually what you do is calculate the derivative of $ f $ and see that $ f^{\prime} \in (a, b) $. Then using the fact that $ a \leq x \leq b $ you deduce that $ | f^{\prime} (x) | <1 $. So I think what is failing is the fact that you have not defined the interval for $ f $.
A: When we write
$$|f(x)-f(y)|=\left|\frac{1}{2\sqrt{\alpha+\xi}}\right||x-y|,\ $$
notice that not only is $\xi$ positive, it's greater than $1$ (for great enough $x$ and $y$, and for $\xi$ is just between $x$ and $y$). So we get
$$\frac{1}{2\sqrt{\alpha+\xi}}<\frac{1}{2\sqrt{\xi}}<\frac{1}{2}.\ $$
That's to say, for each $\alpha>0$, the sequence is convergent.
A: The derivative of $f(x) =\sqrt{\alpha + x}$ is $\frac{1}{2\sqrt{\alpha + x}}$.  If $x \geq 1$, then $$\frac{1}{2\sqrt{\alpha + x}} \leq \frac{1}{2\sqrt{\alpha + 1}} \leq \frac 1 2.$$
This shows that $f(x) = \sqrt{\alpha + x}$ is a contraction on $[1,\infty)$, so there is a fixed point in $[1,\infty)$.
If $x < 1$, since the sequence defined by iterating $f$ is nondecreasing, either it converges to a point below 1 or eventually it ends up in $[1,\infty)$.  In either case it is convergent.
