Area of "polygon" inscribed in circle, whose side lengths form a geometric series $\{v_0, v_n : n\in \Bbb{N}\}$ are points on the circumference of a unit circle, having the following three properties:


*

*The (straight line) distances between neighboring points in the sequence form a convergent geometric series. That is, for all $n>0$, $d(v_n,v_{n+1}) = r\cdot d(v_{n-1},v_n) $ with $r$ a constant ratio for the entire set of verticies, and  $0<r<1$.

*$\lim_{n_to\infty} d(v_n,v_0) = 0$.  That is, the "polygon" formed by these vertices forms a closed shape (in the limit) without having to add a "last" side going back to $v_0$.

*For all $n>0$, the shorter of the two arcs from $v_n$ to $v_{n+1}$ does not contain $v_0$.  That is, the sides go around the circle only once. 
I want to find the area of this "polygon" in terms of the side ratio $r$.  (Although the number of sides needed to close the shape is infinite, one can easily define the area as the limit, as $k$ goes to infinity, of the area formed by the first $k$ sides and an added last side connecting $v_{k-1}$ to $v_0$.)
Several things are immediately clear:  


*

*This area is not defined for $r<\frac12$ because the set of lines cannot from a closed (in the limit) "polygon" unless the sum of the lengths of all the sides other than the first is longer than the first side.  So the function $A(r)$ need only be found on the open interval $(\frac12,1)$.

*For any $r \in( \frac12,1)$, there exists a value of $L_0 \equiv d(v_0,v_1)$ such that the figure exactly closes without overshooting.  $L(r)$ has the property that 
$$ \sum_{n=0}^\infty \sin^{-1}\left(\frac{L}{2}r^n\right) = \pi$$ 

EDIT For $r$ below about $0.6$, the first arc covers more than half the circle, and the area of the polygon does not include that first "super-wedge." In that circumstance, $L(r)$ would have the property that
$$ \sum_{n=1}^\infty \sin^{-1}\left(\frac{L}{2}r^n\right) = \sin^{-1}\left(\frac{L}{2}\right) $$ 
However, my numerical fiddling around indicates that you can't satisfy this, and that makes sense since arcsin is concave upwards.



*

*$\lim_{r\to\frac12} A(r) =0$ and  $\lim_{r\to 1} A(r) =\pi$.  The former represents almost perfectly retracing the first side along a very short arc that is almost a straight line; the latter represents going around the circle in what is almost a regular inscribed polygon.



EDIT The nearly straight line case does not happen; see the edit 
above.

I'm also almost sure that for small positive $\epsilon$,  $A(r)$ has a positive second derivative at $r=\frac12 + \epsilon$, and a  negative second derivative at $r = 1-\epsilon$, and that $A(r)$ has just one point of inflection.
My question, then, is to find $A(r)$, or failing that, to describe some non-trivial properties of $A(r)$.
 A: EDIT.
Function $L(r)$ is implicitly defined by the equation:
$$
\sum_{n=0}^\infty\arcsin\left({1\over2}L(r)\ r^n\right)=\pi
$$
if $r\ge r_0\approx0.602527$, and by the equation
$$
\sum_{n=1}^\infty\arcsin\left({1\over2}L(r)\ r^n\right)=\arcsin\left({1\over2}L(r)\right)
$$
if ${1/2}<r<r_0$. Here's a plot of $L(r)$ obtained by a numerical computation:

If $r\ge r_0$ the polygon is the sum of isosceles triangles $T_n$, having vertex at the center of the circle and base $L_n=L(r)r^n$, so its area is given by
$$
A(r)=\sum_{n=0}^\infty{1\over2}L(r)\ r^n\sqrt{1-{1\over4}L(r)^2r^{2n}}.
$$
If $r<r_0$ the first triangle must be subtracted by the sum of the others, leading to
$$
A(r)=-{1\over2}L(r)\sqrt{1-{1\over4}L(r)^2}+
\sum_{n=1}^\infty{1\over2}L(r)\ r^n\sqrt{1-{1\over4}L(r)^2r^{2n}}.
$$
Here's a plot of $A(r)$:

For a better visualization of the behaviour near $r=1/2$ here's a close-up of the same plot:

A: If a "polygon" with an infinite number of sides of lengths $p^0, p^1, p^2, ...$ is inscribed in a circle, then
$$\lim_{p\to1^-}{(\text{area of circle - area of polygon})=\frac{\pi}{18}}$$
This can be shown by first showing that, as $p\to{1^-}$ the areas of the segments approach a geometric sequence with common ratio $p^3$, then calculating their sum by using the formula for the sum to infinity of a geometric series. (You can use L'Hopital's rule and Maclaurin series to evaluate the limits as $p\to{1^-}$.)
