Convergence of $s_{n+1}=\sqrt{1+s_n}$ Does the sequence $s_{n+1}=\sqrt{1+s_n}$ always converge, no matter what the initial value of $s_1$ is?
Is this sequence always increasing and bounded? I think so, but what's throwing me off is that to find what the sequence converges to, we just solve $s^2=1+s$ to get $s=\frac{1+\sqrt{5}}{2}$.
How can the sequence converge to this number if its inital value is $s_1=3$ for example?

For the case $s_1 = 3$, then $s_2 = 2 < 3 = s_1$ and $f(x) = \sqrt{1+x}\implies f'(x)=\dfrac{1}{2\sqrt{1+x}}> 0\implies f$ is an increasing function $\implies s_n$ is strictly decreasing sequence and is bounded below by $0$ as $s_n > 0, \forall n\ge 1$, hence is convergent to $L$ which is the solution of $L = \sqrt{1+L}\implies L = \dfrac{1+\sqrt{5}}{2}$ as claimed. In general, $s_1 \ge -1$ to begin with. Now if $\dfrac{1-\sqrt{5}}{2} < s_1 < \dfrac{1+\sqrt{5}}{2}\implies s_2 > s_1$ and the sequence is strictly increasing and is bounded above by $\dfrac{1+\sqrt{5}}{2}$ which can be shown by induction on $n \ge 1$. Thus it converges to $L =\dfrac{1+\sqrt{5}}{2}$ again. If $s_1 = \dfrac{1+\sqrt{5}}{2}$, then $s_n = \dfrac{1+\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L =\dfrac{1+\sqrt{5}}{2}$ also. If $s_1 > \dfrac{1+\sqrt{5}}{2}\implies s_n$ is a decreasing sequence as before and is bounded below by $0$ so is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ because $L \ge 0$. If $s_1 = \dfrac{1-\sqrt{5}}{2}\implies s_n = \dfrac{1-\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L = \dfrac{1-\sqrt{5}}{2}$ as $n \to \infty$. Finally, if $-1 \le s_1 < \dfrac{1-\sqrt{5}}{2}\implies s_1 >s_2\implies s_n$ is a strictly decreasing sequence, and is bounded below by $0$ hence is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ since $L \ge 0$. This completes the analysis regarding the possible values of the initial term $s_1$.
