<s>[INST] Prove $u_n$ has the limit given Let $u_n = \sqrt{u_n+1}~ and ~u_1=1$. Prove that $\displaystyle\lim_{n\to\infty} u_n = \frac{1}{2}(1+\sqrt5)$
Here is what I did.
First, to show the sequence converges: 
$u_1 = 1, \space 
u_2 = \sqrt 2 = 2^{(1/2)}, \space 
~u_3 = \sqrt {1 + 2^{(1/2)}} < \sqrt{2*2^{(1/2)}} = 2^{(3/4)} $
therefore 
$u_4 < \sqrt {1 + 2^{(3/4)}}< \sqrt{2*2^{(3/4)}} = 2^{(7/8)}, \space 
~u_5 < \sqrt {1 + 2^{(7/8)}} < \sqrt{2*2^{(7/8)}} = 2^{(15/16)} $
... by weak induction: 
$u_n < 2^((2^{(n-1)} - 1)/2^{(n-1)}) < 2, \space ∀n $
now to find the limit: 
$u_{n+1} = \sqrt {1 + u_n} $
squaring both sides 
$u²_{n+1} = 1 + u_n$ 
so 
$u²_{n+1} - u_n - 1 = 0$ 
let $φ = \lim_{n → ∞} u_n = \lim_{n → ∞} u_{n+1}$ 
and substitute into the equation above: 
$φ² - φ - 1 = 0$ 
now solve for φ with the quadratic formula
 $\displaystyle\lim_{n\to\infty} u_n = \frac{1}{2}(1+\sqrt5)$
 [/INST] $$\left. \begin{align} u_{_{\infty\ +\ 1}} &= \sqrt{u_\infty + 1} \\ \\ \infty + 1 &= \infty \end{align}\ \right\} \iff u_\infty = \sqrt{u_\infty + 1} \iff u_\infty^2 = u_\infty + 1 \iff u_\infty^2 - u_\infty - 1 = 0$$ $$u_\infty = \frac{1\pm\sqrt5}2$$ Obviously, since a radical is always positive, the only possible solution is $u_\infty = \frac{1\ +\ \sqrt5}2 = \varphi > 0$ .
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