$a, z_1 \gt 0$ and $z_{n + 1} = (a + z_n)^{\frac 1 2}$ then $(z_n)$ is monotone and bounded?

Question

Having trouble with the following exercise in Bartle's Elements of Real Analysis.

Let $a, z_1 \gt 0$. Define $z_{n + 1} = (a + z_n)^{\frac 1 2}$ for $n \in \Bbb N$. Show that $(z_n)$ converges.

Now the hints section suggests $(z_n)$ is monotone and bounded. I disagree. If you pick $a = 100,00$ and $z_1 = 1$ this is apparent. But I think the sequence is monotone for $n \gt 1$. And I tried the following ugly argument (not even sure if I'm right):

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If $z_1 \le (z_1 + a)^{\frac 1 2}$ then $(z_1 + a)^{\frac 1 2} \le \left({(z_1 + a)^{\frac 1 2} + a }\right)^{\frac 1 2} \implies z_2 \le z_3$. If $z_n \le z_{n + 1}$ for $n \gt 1$ then $(z_n + a )^{\frac 1 2} \le (z_{n+ 1} + a )^{\frac 1 2} \implies z_{n + 1} \le z_{n + 2} \implies z_n \le z_{n + 1} \;\; \forall n \in \Bbb N \setminus \{1\}$. So the sequence is monotone increasing. OR we have that $z_1 \gt (z_1 + a)^{\frac 1 2}$ and using pretty much the same argument we can show that the sequence is monotone decreasing. For the second case the bound is $0$.

But I am unable to find a bound when the sequence is monotone increasing.

Answer 1

Using only the fact that the function $x\mapsto\sqrt{a+x}$ is increasing and that the sequence is defined by $z_1\geqslant0$ and $z_{n+1}=u(z_n)$, one sees that:

• If $z_1\lt z_2$, the sequence $(z_n)$ is increasing
• If $z_1\gt z_2$, the sequence $(z_n)$ is decreasing
• If $z_1=z_2$, the sequence $(z_n)$ is constant

These three conditions are easily converted in terms of $z_1$ alone, as $$ z_1\lt z^*,\qquad z_1\gt z^*,\qquad z_1=z^*, $$ respectively, where $$ z^*=\frac{1+\sqrt{1+4a}}2. $$