If $a>0$ and $z_{1}>0$ define $z_{n+1}=\sqrt{a+z_{n}}$ show that it converges and find the limit

Question

I have attempted this problem and came up with the following method to solve it

One can show that the sequence $z_{n+1}=\sqrt{a+z_{n}}$ is monotone I assumed it's increasing as a case study , therefore relying on the monotone convergence theorem

I tried to estimate a possible candidate for a bound, assuming $(z_{n})$ converges we get

$$\lim_{n}z_{n+1}=\lim_{n}\sqrt{a+z_{n}}\implies z=\sqrt{a+z}$$

therefore

$$z^2-z-a=0\implies z=\frac{1\pm\sqrt{1+4a}}{2}=\frac{1}{2}\pm \sqrt{\frac{1}{4}+a}$$

If the sequence converges choose $z=\frac{1}{2}+\sqrt{1/4+a}$ this would be the sequences supremum and $\frac{1}{2}\pm \sqrt{1/4+a}\leq 4+a$

therefore I made the claim that $z_{n}\leq a+4, \; \forall n\in\mathbb{N}$, and I tried to prove it as follows

assume there exists $n_{0}\in\mathbb{N}$ such that $z_{n_{0}}>a+4$ then

$$S=\{n\in\mathbb{N}:z_{n}>a+4\}\neq \emptyset$$

thus by the well ordering principle it attains a least element say $m$ but then

$$\begin{align}z_{m}&>a+4 \\ \sqrt{a+z_{m-1}}&>a+4 \\ a+z_{m-1}&>a^2+8a+16 \\ z_{m-1}>(a+4)+(a^{2}+6a+12)&>a+4 \end{align}$$

This would contradict the choose of $m$ and create an infinite descent therefore, $(z_{n})$ is a monotone and bounded sequence thus it's convergent and it's limit is the supremum suggested above

"WLOG I assumed it's increasing" $;-;$ You'll have to justify the "WLOG" claim since it's not obvious here. — dxiv, May 13 '23 at 22:40
@dxiv my bad I meant to write I considered the case where $(z_{n})$ is increasing — Eulerroid, May 13 '23 at 22:41
Does this answer your question? Let $a>0$ and $x_1 > 0$ and $x_{n+1} = \sqrt{a + x_n}$ for $n \in \mathbb{N}$. Show that ${x_n}_{n\ge 1}$ converges - found using an Approach0 search. Note the search results include multiple other duplicates, with some of these closed as duplicates of other questions and/or have links to them, ... — John Omielan, May 14 '23 at 03:29
(cont.) e.g., Show that the following inductively defined sequence converges and find its limit., Convergence and limit of a recursive sequence $x_{n+1}=\sqrt{p+x_n}$, Monotone series increases and decreases depending on first term. What is the limit?, etc. — John Omielan, May 14 '23 at 03:32

score 2 · Accepted Answer · answered May 14 '23 at 00:27

A DIRECT PROOF OF CONVERGENCE

First note that all $z_n$ are positive.

Let $t=\frac{1+\sqrt{1+4a}}{2}$, then $t>1$ and $t^2=a+t$.

Then $z_{n+1}^2-t^2=a+z_n-(a+t)=z_n-t$. Therefore

$$ |z_{n+1}-t|=\frac{|z_n-t|}{|z_{n+1}+t|}<\frac{|z_n-t|}{t}.$$

And therefore $\{z_n\}$ tends to $t$.

COMMENTS ON YOUR SOLUTION

You are correct in stating that the sequence is monotone but this does need proving. (The proof is similar to the one above, replacing $t$ by $z_n)$.

Your argument that $a+4$ is an upper bound is clever but is not strictly correct since the descent need not be infinite but could end with $z_1>a+4$. Since $z_1$ is arbitrary this could happen and in that case you would have to show that the sequence is monotonic decreasing.

I hope these comments are of use.

If $a>0$ and $z_{1}>0$ define $z_{n+1}=\sqrt{a+z_{n}}$ show that it converges and find the limit

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