Prove the limit as $n$ tends to infinity for a bounded monotonic sequence $u_{n+1} = \sqrt{u_n + 1}$.

Question

Saw this problem in my textbook the other day, and since I'm not very familiar with bounded monotic sequences, I've been struggling with it.

If $u_{n+1} = \sqrt{u_n + 1}$ and $u_1 = 1$, prove that $\lim_{n \to \infty} u = 1/2(1 + \sqrt{5})$.

Since I'm not that good with sequences yet, and I'm new on this site, I'd like to request that whoever's answering try to simplify things a bit more. Thanks!

Answer 1

HINT

We need to proceed as follow

• $u_{n+1} = \sqrt{u_n + 1}\geq u_n > 0\implies $ $u_n$ is positive and strictly increasing.
• $u_n$ is bounded.

(both to be proved rigorously by induction)

then by monotone sequences theorem the limit exists, then assume $x_n\to L$ and we have

$$L=\sqrt{L+1}\implies L^2-L-1=0 \implies L=\frac{1+\sqrt 5}{2}$$

Answer 2

1. Show by induction, that $(u_n)$ is increasing.

2. Show by induction, that $u_n \leq 2$ for all $n$.

It follows that $(u_n)$ is convergent.

Let $u= \lim\ u_n$. Then $u = \sqrt{u + 1}$.

Can you proceed ?