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Here is Prob. 6, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $a > 0$ and let $z_1 > 0$. Define $x_{n+1} \colon= \sqrt{ a + z_n } $ for $n \in \mathbb{N}$. Show that the sequence $\left( z_n \right)_{n \in \mathbb{N} }$ converges, and find the limit.

My Attempt:

We have the following two cases, according to whether $z_1 \geq \frac{ 1 + \sqrt{ 1 + 4a } }{2}$ or $z_1 < \frac{ 1 + \sqrt{ 1 = 4a } }{2}$:

Case 1. When $z_1 \geq \frac{ 1 + \sqrt{ 1 + 4a } }{2}$, and as $$ \frac{ 1 + \sqrt{ 1 + 4a } }{2} > \frac{ 1 - \sqrt{ 1 + 4a } }{2}, $$ we have $$ z_1 > \frac{ 1 - \sqrt{ 1 + 4a } }{2}$$ also and then $$ z_1^2 - z_1 - a = \left( z_1 - \frac{ 1 - \sqrt{ 1 + 4a } }{2} \right) \left( z_1 - \frac{ 1 + \sqrt{ 1 + 4a } }{2} \right) \geq 0, $$ and hence $$ z_1^2 \geq a + z_1, $$ which implies (since both of $z_1$ and $a + z_1$ are positive) that $$ z_1 \geq \sqrt{ a+ z_1} = z_2. $$

Now suppose that, for some $k \in \mathbb{N}$, we have $z_k \geq z_{k+1}$. Then we also have $$ z_{k+1} = \sqrt{ a+z_k} \geq \sqrt{ a + z_{k+1} } = z_{k+2}. $$ Therefore by induction we can conclude that $$ z_n \geq z_{n+1} \ \mbox{ for all } n \in \mathbb{N}. \tag{1} $$

We also note that, as $$z_1 \geq \frac{ 1 + \sqrt{1 + 4a } }{2} > \frac{ \sqrt{4a } }{2} = \sqrt{a}, $$ so we have $$ z_1 > \sqrt{a} > 0, $$ and thence $$ z_2 = \sqrt{ a + z_1} > \sqrt{a} > 0. $$ [This last step can be treated as redundant.]

Suppose that, for some $k \in \mathbb{N}$, we have $$ z_{k} > \sqrt{a}. $$ Then we also have $$ z_{k+1} = \sqrt{ a + z_{k} } > \sqrt{ a + a } = \sqrt{2a} > \sqrt{a}. $$ Therefore, by induction we can conclude that $$ z_{n} > \sqrt{a} \ \mbox{ for all } n \in \mathbb{N}. \tag{2} $$ Thus our sequence is bounded from below.

From (1) and (2) it is evident that our sequence $\left( z_n \right)_{n \in \mathbb{N} } $, being monotonically decreasing and bounded below, is convergent. Let us put $$ z \colon= \lim_{n \to \infty} z_n. $$ Then we obtain $$ z = \sqrt{a + z}, \tag{3} $$ which implies that $$ z^2 = a+z, $$ and hence that $$ z^2 - z - a = 0, $$ and so $$ z = \frac{ 1 \pm \sqrt{ 1 + 4a } }{2}. $$ But from (2) it also follows that $$ \lim_{n \to \infty} z_n \geq \sqrt{ a} > 0. $$ Hence we cannot have $z = \frac{1 - \sqrt{ 1+4a} }{2}$. Hence we must have $$ z = \frac{ 1 + \sqrt{ 1 + 4a } }{2} . \tag{A}$$

Case 2. When $0 < z_1 < \frac{ 1 + \sqrt{ 1+4a } }{2}$, and as $$ z_1 > 0 > \frac{ 1 - \sqrt{ 1 + 4a } }{2}$$ also holds, so we have
$$ z_1^2 - z_1 - a = \left( z_1 - \frac{ 1 - \sqrt{ 1 + 4a } }{2} \right) \left( z_1 - \frac{ 1 + \sqrt{ 1 + 4a } }{2} \right) < 0, $$ and hence $$ z_1^2 < a + z_1, $$ which implies (since both of $z_1$ and $a + z_1$ are positive) that $$ z_1 < \sqrt{ a+ z_1} = z_2. $$

Now suppose that, for some $k \in \mathbb{N}$, we have $z_k < z_{k+1}$. Then we also have $$ z_{k+1} = \sqrt{ a+z_k} < \sqrt{ a + z_{k+1} } = z_{k+2}. $$ Therefore by induction we can conclude that $$ z_n < z_{n+1} \ \mbox{ for all } n \in \mathbb{N}. \tag{4} $$

Moreover, as $$ 0 < z_1 < \frac{ 1 + \sqrt{ 1+4a } }{2}, $$ and as $$ \frac{ 1 + \sqrt{ 1+4a } }{2} < \frac{ 1 + \sqrt{ 1 + 4 \sqrt{a} +4a } }{2} = \frac{ 1 + \sqrt{ \left( 1 + 2 \sqrt{a} \right)^2 } }{2} = \frac{ 1 + (1 + 2 \sqrt{a} ) }{2} = 1 + \sqrt{a}, $$ so we also obtain $$ z_1 < 1 + \sqrt{ a}, $$ and thence $$ z_2 = \sqrt{ a+ z_1 } < \sqrt{ a + 1 + \sqrt{ a } } < \sqrt{ a + 1 + 2 \sqrt{a} } = \sqrt{ \left( 1 + \sqrt{a} \right)^2 } = 1 + \sqrt{a}. $$ [This last chain of inequalities that we have obtained may be treated as redundant.]

Now suppose that, for some $k \in \mathbb{N}$, we have $$ z_k < 1 + \sqrt{a}. $$ Then we obtain $$ z_{k+1} = \sqrt{ a + z_k} < \sqrt{ a + 1 + \sqrt{a} } < \sqrt{ a + 1 + 2 \sqrt{a} } = \sqrt{ \left( 1 + \sqrt{a} \right)^2 } = 1 + \sqrt{a}. $$ Therefore by induction we can conclude that $$ z_n < 1 + \sqrt{a} \ \mbox{ for all } n \in \mathbb{N}. \tag{5} $$

From (4) and (5) we it is evident that our sequence $\left( z_n \right)_{n \in \mathbb{N} }$, being monotonically increasing and bounded from above, is convergent.

Now as in Case 1, if we put $$ z \colon= \lim_{n \to \infty} z_n, $$ then from (5) we can gaurantee that our $z$ must satisfy (3) along with $$ 0 \leq z \leq 1 + \sqrt{a}. $$ Then, using the same procedure as in Case 1, we obtain $$ z = \frac{1 + \sqrt{ 1+4a } }{2}. \tag{B} $$

Finally, from (A) and (B) we conclude that, in either case, we have $$ \lim_{n \to \infty } z_n = \frac{ 1 + \sqrt{ 1 + 4a } }{2}. $$

Is my proof correct and clear enough in each and every step of it? Or, are there any loopholes in terms of its logic or understandability to an elementary-level learner of real analysis?

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