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!
 A: You have
$$
x_{n+1} = \sqrt{3+x_n}
$$
so that
$$
x_{n+1}^2 = 3+x_n.
$$


*

*Assuming the sequence converges to a limit $\ell\in\mathbb{R}$, then $\ell$ must satisfy (by continuity)
$$
\ell^2 = 3+\ell
$$
hence the only possible values for $\ell$ are therefore $\frac{1\pm\sqrt{13}}{2}$. Since the limit of a positive sequence has to be non-negative, if the sequence has a limit then this limit is $\ell=\frac{1+\sqrt{13}}{2}$.

*Now, even if this does not seem to be requested by the question, let us also prove there is such a limit (that is, that the sequence converges).


*

*We can easily see that $x_n \geq x_{n-1} \geq \sqrt{3}$ for all $n\geq 2$. (e.g., by induction, from the recurrence relation: this is true for $x_2 \geq \sqrt{3} = x_1$, and then if $x_n\geq x_{n-1}$ then $3+x_n \geq 3+x_{n-1}$ by induction hypothesis)).

*We can show that it is bounded above by, say, $3$. Again, by induction: if $x_n \leq 3$, then $x_{n+1} = \sqrt{3+x_n} \leq \sqrt{3+3} < 3$.

*A bounded non-decreasing sequence converges.
A: 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!
A: You came up with a recurrence relation; that is, $x_{n}$ as a function of $x_{n-1}$:
$$
x_n = \sqrt{x_{n-1} + 3}
$$
If a sequence converges to something, that something must be a fixed point of the function that takes $x_{n-1}$ to $x_n$. Do you see why?
So, we say "Well, if $x_{n-1}$ is to equal $x_n$, then any values $L$ at which they are equal must be the solutions of the following:"
$$
\begin{aligned}
L &= \sqrt{L + 3}\\
L^2 &= L + 3\\
L^2 - L - 3 &= 0\\
\end{aligned}
$$
So, using the quadratic formula, we have:
$$
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}
$$
You can probably guess that the limit will be the positive one, $\frac{1 + \sqrt{13}}{2}$. In order to conclude this, you should show that the function is increasing. Or, actually, you could just show that every term in the sequence is positive (they're sums of square roots of positive things).
(But note that the fact that $\frac{1 - \sqrt{13}}{2}$ is a solution shows that you need to know the initial conditions to determine the behavior of this sequence. If $x_1$ were $\frac{1 - \sqrt{13}}{2}$, then the sequence would just stay there forever!
But all other starting conditions will lead to $\frac{1 + \sqrt{13}}{2}$. [Unless they're too negative, i.e. $x_1 = -10$ will leave you with $x_2 = \sqrt{-7}$]. It's an attracting fixed point, while $\frac{1 - \sqrt{13}}{2}$ is a repelling fixed point.)
