<s>[INST] Prove $\sqrt{s_n+1} = \frac{1}{2}(1+\sqrt{5})$ This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
 [/INST] Argue by induction: 


*

*The terms of the sequence are all positive. 

*They are all bounded above (by $3$): The base case is clear, and if $s_n<3$, then $s_{n+1}=\sqrt{s_n+1}<\sqrt{3+1}=2<3$. 

*The sequence is increasing: $s_2=\sqrt2>1=s_1$, and if $s_n<s_{n+1}$, then $s_n+1<s_{n+1}+1$, so $s_{n+1}=\sqrt{s_n+1}<\sqrt{s_{n+1}+1}=s_{n+2}$.


It follows that the sequence converges (to its supremum). Call its limit $L$, so $L=\lim_n s_{n+1}=\lim_n\sqrt{s_n+1}=\sqrt{L+1}$ (by continuity of $f(x)=\sqrt{x+1}$).
Solving the equation $L=\sqrt{L+1}$ gives us that $L^2-L-1=0$, so $L=\frac12(1\pm\sqrt5)$, and the sign must be $+$ rather than $-$ since the terms of the sequence are all positive, so also their limit $L$ is non-negative, $L\ge0$.
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