$\lfloor a_n \rfloor$=? where $a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$ $\lfloor a_n \rfloor$=?, where $$a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$$
and the number of radicals is equal to n.
$\lfloor a_1]=\lfloor \sqrt{2005} \rfloor=44$, $\lfloor a_2 \rfloor=\lfloor \sqrt{2005+\sqrt{2005}} \rfloor=45$,  $\lfloor a_3 \rfloor=\lfloor \sqrt{2005+\sqrt{2005+\sqrt{2005}}} \rfloor=45$ and so on,
but at some point becomes $\lfloor a_k \rfloor=46$ and then $47$ and so on.
So we can say that $a_n$ depends on the value of $n$, which in my opinion has no rule.
In order, how it should be defined $\lfloor a_n \rfloor$=?
Thank you!
 A: The sequence is an increasing one. First, we show that it is bounded above by 46 which is a simple induction exercise.
Since an increasing bounded above sequence is convergent our sequence has a finite limit.
Assuming the limit is L we have  $$ L=\sqrt {2005+L}$$
Solving for L we get $$L=45.28001707...$$
Thus for $n\ge 2$,  $$\lfloor a_n \rfloor=45$$
A: $\def\e{\varepsilon}$Here we bound the sequence by examining the sequence of successive differences.
Let $x=2005$.
We have the recursion
\begin{align*}
a_0 &= 0 \\
a_n &= \sqrt{x+a_{n-1}},\quad n\ge1.
\end{align*}
Note that $a_n\ge0$ for $n\ge 0$.
Let $d_n = a_n-a_{n-1}$ for $n\ge 1$.
One can verify that
\begin{align*}
d_n &= \frac{d_{n-1}}{\sqrt{x+a_{n-1}}+\sqrt{x+a_{n-2}}},
\quad n\ge 2.
\end{align*}
Since $d_2 = \sqrt{x+\sqrt x}-\sqrt{x} > 0$, this implies that $d_n > 0$ for $n\ge 2$.
Note that $d_1=a_1-a_0 = \sqrt x>0$.
Thus, $a_n$ is increasing and so
$$a_n > a_{n-1} > \cdots > a_0 = 0.$$
Thus, $d_n < \e d_{n-1}$ for $n\ge 2$, where $\e = 1/(2\sqrt{x}).$
More generally,
$$d_n < \e d_{n-1} < \e^2 d_{n-2} < \cdots < \e^{n-2}d_2 < \e^{n-1}d_1.$$
The last inequality follows since $d_2<\e d_1$.
Thus,
\begin{align*}
a_n &= a_0 + \sum_{k=1}^n d_k 
< d_1\sum_{k=1}^n \e^{k-1} 
< \sqrt x \sum_{k=1}^\infty \e^{k-1} 
=  \frac{\sqrt{x}}{1-\e} 
= \frac{2x}{2\sqrt{x}-1}.
% &= 45.2829\ldots
\end{align*}
Therefore, $a_n < 45.2829\ldots$.
(This technique can be extended to find accurate closed form approximations for $a_n$.)
Since
\begin{align*}
a_0 &= 0 \\
a_1 &= \sqrt{x} = 44.7772 \\
a_2 &= \sqrt{x+\sqrt x} = 45.2745
\end{align*}
we find
$$\lfloor a_n \rfloor = \begin{cases}
0, & n = 0 \\
44, & n = 1 \\
45, & n\ge 2
\end{cases}$$
agreeing with the result of @MohammadRiazi-Kermani.
