<s>[INST] Real Analysis proof, show there is a limit and find it. Assume$${ x }_{ 1 }=\sqrt { 3 } ,{ x }_{ 2 }=\sqrt { 3+\sqrt { 3 }  } ,..,{ x }_{ n }=\sqrt { 3+\sqrt { 3+..+\sqrt { 3 }  }  } $$this sequence has a limit and find the limit.
I know I need to show that is is bounded and then show $x_{ n }<{ x }_{ n+1 }$ before finding the limit but have no idea how to go about it. Please help!
 [/INST] We have the recurrence relationship
$$x_{n+1}=\sqrt{3+x_n}$$
with $x_1=\sqrt{3}$.  
We propose that $\lim_{n\to \infty}x_n=\frac{1+\sqrt{13}}{2}$.  

First note that $x_n>0$.  
Second, we observe that if $x_{n}<\frac{1+\sqrt{13}}{2}$, then $x_{n+1}<\frac{1+\sqrt{13}}{2}$ also.  
Therefore, for $x_1=\sqrt3<\frac{1+\sqrt{13}}{2}$, the sequence $x_n$ is bounded above by $\frac{1+\sqrt{13}}{2}$.

Third, note that for $0<x_n<\frac{1+\sqrt{13}}{2}$
$$\begin{align}
x_{n+1}-x_n&=\sqrt{3+x_n}-x_n\\\\
&=-\frac{x_n^2-x_n-3}{\sqrt{3+x_n}+x_n}\\\\
&=-\frac{\left(x_n-\frac{1+\sqrt{13}}{2}\right)\left(x_n-\frac{1-\sqrt{13}}{2}\right)}{\sqrt{3+x_n}+x_n}\\\\
&>0
\end{align}$$
Therefore, the sequence $x_n$ is increasing monotonically and bounded above.  Therefore, the limit $\lim_{n\to \infty}x_n$ exists and is finite.

Let $L$ represent the limit.  Then, 
$$\begin{align}
L&=\lim_{n\to \infty}x_{n+1}\\\\
&=\lim{n\to \infty}\sqrt{3+x_n}\\\\
&=\lim{n\to \infty}\sqrt{3+\lim{n\to \infty}x_n}\\\\
&=\sqrt{3+L} \tag 1
\end{align}$$
whereupon equating the left-hand and right-hand sides of $(1)$ reveal that $L=\frac{1+\sqrt{13}}{2}$ as was proposed!
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