1. ## Sequence Proof

Given the following sequence:

$x_0 = 1, x_1 = \sqrt{3+1}, x_2 = \sqrt{3+\sqrt{4}}, x_3 = \sqrt{3+\sqrt{5}},$

$x_4 = \sqrt{3+\sqrt{3+\sqrt{5}}}, x_5 = \sqrt{3+\sqrt{3+\sqrt{3+\sqrt{5}}}} \ldots$

Prove the above sequence converges and determine the limit.

2. I haven't found a proof of convergence yet, but the limit (assuming it converges) is $\frac{1+\sqrt{13}}{2}$

the sequence $x_{n+1} = 3 + \sqrt{x_n}$ has a stable point of equilibrium when $x=3+\sqrt{x}$
Solving this for $\sqrt{x}$ gives $\frac{1\pm\sqrt{13}}{2}$
Since the square root of this sequence is the sequence we want to find the limit of, these are the equilibrium points of our sequence. Since it is readily apparent that we initially move toward $\frac{1+\sqrt{13}}{2}$, we can tell that if the sequence converges, it will converge to this point.

I will post again if I find a proof it converges.

3. From n=3 on the sequence is defined recursively: $n \geqslant 4,\;x_n = \sqrt {3 + x_{n - 1} }$.
Show that this sequence is increasing and bounded above.
That will prove convergence.

4. How do I show that a recursively defined function is increasing and bounded ?

5. Originally Posted by Auxiliary
How do I show that a recursively defined function is increasing and bounded
The following should be clear to you.
$\begin{array}{l}
x_2 < x_3 < 9 \\
\sqrt {x_2 } < \sqrt {x_3 } < 3 \\
3 + \sqrt {x_2 } < 3 + \sqrt {x_3 } < 6 \\
\underbrace {\sqrt {3 + \sqrt {x_2 } } }_{x_3 } < \underbrace {\sqrt {3 + \sqrt {x_3 } } }_{x_4 } < \sqrt 6 \\
x_3 < x_4 < 9 \\
\end{array}$

Using that pattern and induction prove it.

6. Ok, so in general:

$x_{n+1} = \sqrt{3+\sqrt{x_n}} < \sqrt{3+3} < \sqrt{3+6} = 3$

And so I've shown that it is bounded. Now how do I show it's monotone?

7. Originally Posted by Auxiliary
And so I've shown that it is bounded. Now how do I show it's monotone?
How did I show $x_3 < x_4$?
You do it the same say using induction.