real analysis - How do I prove the sequence $x_{n+1}=sqrt{alpha +x_n}$, with $x_0=sqrt alpha$ converges? ( boundedness?)

I need to "study the limit behavior of the following sequence" and then compute the limit. I can compute the limit and prove the monotonicity, but I am having trouble proving boundedness. I tried to understand from other similar posts how this could be done, but since I didn't understand I'm asking it as a new question.
The sequence is $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha>0$.

It seems to be monotonically increasing.

Proof:

It can be shown by induction that the sequence is monotonically increasing.

Claim that $x_{n+1}\geq x_n$
$$x_1=\sqrt{\alpha+\sqrt\alpha}>\sqrt\alpha=x_0$$
$x_{n+1}\implies x_{n+2}$

$$x_{n+1}=\sqrt{\alpha+x_n}<\sqrt{\alpha+x_{n+1}}=\sqrt{\alpha+\sqrt{\alpha+\sqrt\alpha}}=x_{n+2}$$
Boundedness:

Because the sequence is monotonically increasing and $x_o=\sqrt\alpha$, it is bounded below by $\sqrt\alpha$. Now comes the part where I have to prove that the sequence is bounded above. The problem is that I don't know how to start, so if anyone could give me a bit of a push I'd be grateful.

The limit:

I know that $\lim\ x_n=\lim \ x_{n+1}$, so letting $\lim\ x_n=x$ then
$$x=\sqrt{\alpha+x}$$ and solving would give $x=\frac{1+\sqrt{1+4\alpha}}{2}$.

May somebody please confirm my proof thus far, and help me prove the upper bound?
Thanks!

Answer

Instead of just proving the boundedness, let me show you the logic to construct the proof.

Step 1 guess the answer.

In this step, you don't need to be rigorous. You just use whatever intuition to guide you for a potential candidate of answer.

Let's say you want an upper bound $L$.

What does it mean? It means you want an $L$ such that for any $x \in (0,L]$, $\sqrt{\alpha+x} \le L$.

Since $\sqrt{\alpha+x}$ is an increasing function in $x$, you want

$$\begin{align}&\sqrt{\alpha+L} \le L\\
\iff &\alpha + L \le L^2\\
\iff &\alpha \le L^2 - L\\

\iff &\alpha + \frac14 \le (L - \frac12)^2\\
\implies &\sqrt{\alpha + \frac14} + \frac12 \le L\tag{*}
\end{align}$$
Notice $\sqrt{\alpha + \frac14}$ is bounded above by $\sqrt{\alpha} + \sqrt{\frac14} = \sqrt{\alpha} + \frac12$. This means if you have chosen $L$ such that $\sqrt{\alpha} + 1 \le L$, $(*)$ will be satisfied.

Step 2 verify you answer.

Once you have a potential candidate for an answer, you need to verify
it. If your logic of finding the candidate is reasonable. You can

usually reverse the steps to prove it is indeed what you want.

Let $L = \sqrt{\alpha}+1$, for any $0 \le x \le L$, we have:

$$\sqrt{\alpha + x} \le \sqrt{\alpha + L} = \sqrt{\alpha+ \sqrt{\alpha}+1} < \sqrt{\alpha + 2\sqrt{\alpha}+1} = \sqrt{\alpha} + 1 = L$$

This means if you start with a $x_n \le L$, you will get $x_{n+1} = \sqrt{\alpha + x_{n}} \le L$.

Since $x_0 \le L$, the boundedness of all $x_n$ is proved.

BTW, the rest of your steps look fine.