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 $\frac{1}{2}(1+\sqrt{5})$?
 A: 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, $L\ge0$.
A: Consider applying the Contraction Mapping Theorem. Just check that the conditions for the theorem are satisfied (I'll leave those details to you).
We have $s_{n+1} = \sqrt{s_n + 1}$. Take $F: x \to \sqrt{x +1}$. By the Contraction Mapping Theorem we can conclude that $F$ has a unique fixed point $s$.
This gives us:
\begin{align*}
   \lim s_{n+1} =& \lim\sqrt{s_n +1} \\
   s =& \sqrt{s+1}  \\
   s^2 - s - 1 =& 0
\end{align*}
Then by the quadratic formula we have that our unique fixed point $s = \dfrac{1+\sqrt5}{2}$.
