Find the value of $n$ for these circles. 
Circle $B$, which has radius $2008$, is tangent to horizontal line $A$ at point $P$. Circle $C_1$ has radius $1$ and is tangent both to circle $B$ and to line $A$ at a point to the right of point $P$. Circle $C_2$ has radius larger than $1$ and is tangent to line $A$ and both circles $B$ and $C_1$. For $n > 1$, circle $C_n$ is tangent to line $A$ and both circles $B$ and $C_{n–1}$. Find the largest value of $n$ such that this sequence of circles can be constructed through circle $C_n$ where the $n$ circles are all tangent to line $A$ at points to the right of $P$.


I tried forming series of equations using Pythagoras theorem and the radii $r_i$ of the circles:
$r_2=r_1*\frac{2008+r_1}{2008-r_1}$ and $r_3=r_2*\frac{2008+r_2}{2008-r_2}$ etc.
but I'm not sure what the limiting case will for $n$ and how to use it for finding the last equation. Can someone suggest an alternate approach or a limiting case?
Any help is appreciated.
 A: Let $R=2008$, $r_i$ is radius of $i$-th circle, $x_i$ is distance from $P$ to tangent point of $i$-th circle with line $A$.
Then from Pythagoras theorem:
$$\begin{aligned}
(R+r_i)^2=(R-r_i)^2+x_i^2 \\
(R+r_{i+1})^2=(R-r_{i+1})^2+x_{i+1}^2 \\
(r_i+r_{i+1})^2=(r_{i+1}-r_i)^2+(x_{i+1}-x_i)^2
\end{aligned}$$
Solving for $r_{i+1}$ in terms of $r_i$ gives
$$r_{i+1}=\frac{Rr_i}{\left(\sqrt{R}-\sqrt{r_i}\right)^2}$$
Last circle has radius greater than or equal to $R$. Sequential numerical calculation of all $r_i$ starting from $r_1=1$ gives that $r_{44}<R$, $r_{45} > R$. Then maximum $n$ is 45. Next circle may be tangent to $A$ at point to the left of $P$, but this is prohibited by problem statement.

Deleted after reading comments:
I don't know if there is simple way to check at what $i$ radius $r_i$ becomes greater than $R$.
Added after reading comments:
$$r_{i+1}=\frac{Rr_i}{\left(\sqrt{R}-\sqrt{r_i}\right)^2}\Rightarrow
\frac1{\sqrt{r_{i+1}}}=\frac{\sqrt{R}-\sqrt{r_i}}{\sqrt{Rr_i}}=
\frac1{\sqrt{r_i}}-\frac1{\sqrt{R}}$$
$$\frac1{\sqrt{r_n}}=\frac1{\sqrt{r_1}}-\frac{n-1}{\sqrt{R}}$$
$$r_n\geq R \Rightarrow \frac1{\sqrt{r_n}}\leq \frac1{\sqrt{R}}
\Rightarrow\frac1{\sqrt{r_1}}\leq\frac{n}{\sqrt{R}} \Rightarrow
n\geq \sqrt{\frac{R}{r_1}}$$
$$n=\left\lceil \sqrt{\frac{R}{r_1}} \right\rceil=45$$
