Solar Sail and GR In Newtonian gravity a theoretical reflective sail could be made such that gravity pulling it down toward the star is compensated by 'light pressure' coming from the same star.
In both cases the force drop like $\frac{1}{r^2}$ 
therefore the forces will compensate at any distance.
If we introduce GR we have both a correction to the laws of gravity, and a redshift that changes the outward force.
So my question Is, at the limit $ \frac{r}{r_s}\rightarrow 0 $ Would the sail be attracted or repelled by the star?
 A: First, if you have an extremely large sail near a very powerful star, the 'photon pressure' on the sail can be greater than the pull due to gravity. Even though both forces decrease as $r^{-2}$ this doesn't necessarily mean that the two forces will always cancel, one of them can be always greater than the other (in the case of a 'light sail' this depends on the size of the sail).
In the $r \to \infty$ limit, far away from the star, you recover normal Newtonian gravity and the redshift becomes negligible. It is interesting to include the effects of redshift and GR on the efficiency of the solar sail - as you point out - but these effects only become very relevant when you would be close to the star. As you move further away from the star, these relativistic corrections become negligible.
That being said, let's consider the case including the relativistic effects: for a particle moving in a Schwarzschild spacetime
$$
c^{2}d\tau^{2} = c^2 \left( 1-\frac{\text{r}_s}{r} \right) dt^{2} - \left( 1-\frac{\text{r}_s}{r} \right)^{-1} dr^{2} - r^{2} d\theta^{2} - r^{2} \sin^{2} \theta d\phi^{2}
$$
The equation of motion for a free massive particle moving along a fixed radial trajectory (ignoring tangential motion along the $\theta$ and $\phi$ directions) in this Schwarzschild spacetime is then given by
$$
0 = \frac{d^2 r}{d\tau^2}  -\frac{\text{r}_s}{2 r^2-2 r \text{r}_s} \left( \frac{dr}{d\tau} \right)^2 + \frac{c^2 \text{r}_s (r-\text{r}_s)}{2 r^3} \left( \frac{dt}{d\tau} \right)^2
$$
Now, if we have an ideal solar sail (with 100% reflection) that has a surface area that covers $A$ steradians at our star’s surface (located at $\text{r}_*$. Let us neglect the internal processes in the star; say that we know it’s surface luminosity $L_*$ (the amount of energy it emits at its surface). When the solar sail is located at $\text{r}_*$ (close to the surface of the star) it will receive a momentum $p = \frac{L_* A}{4 \pi r_*^2 c}$. 
As we get further away from the star the energy the solar sail will receive decreases due to relativistic redshift, giving us that at a radius $r$ the effective luminosity is given by
$$
L(r) = \frac{L_* }{\sqrt{1-\frac{\text{r}_s}{r} }}
$$
The momentum our solar sail then picks up at this distance is
$$
p(r) = \frac{L(r) A}{4 \pi r^2 c} = \frac{L_* A}{4\pi r^2 c \sqrt{1-\frac{\text{r}_s}{r} }}
$$
We can now include the momentum that our solar sail receives as a function of $r$ in the geodesic equation that we wrote down earlier, where we now include the force acting on our solar sail on the left hand side 
$$ f^i = m \left( \frac{d^2 x^i}{dt^2} + \Gamma^i_{jk} \frac{dx^j}{dt}\frac{dx^k}{dt} \right)$$
Where now 
$$ f^i = \frac{dp}{d\tau} = \frac{dr}{d\tau} \frac{d}{dr} p(r) = \frac{A L_* (3 \text{r}_s-4 r)}{8 \pi  c r^3 (r-\text{r}_s) \sqrt{1-\frac{\text{r}_s}{r}}} \frac{dr}{d\tau} $$
The complete equation of motion for our solar sail now becomes 
$$
\frac{A L_* (3 \text{r}_s-4 r)}{8 \pi  c r^3 (r-\text{r}_s) \sqrt{1-\frac{\text{r}_s}{r}}} \frac{dr}{d\tau} = \frac{d^2 r}{d\tau^2}  -\frac{\text{r}_s}{2 r^2-2 r \text{r}_s} \left( \frac{dr}{d\tau} \right)^2 + \frac{c^2 \text{r}_s (r-\text{r}_s)}{2 r^3} \left( \frac{dt}{d\tau} \right)^2
$$
Had we not included the relativistic effects, the differential equation would instead have been of the form
$$ \frac{A L_*}{2 c \pi  r^3} = \frac{d^2 r}{d\tau^2}$$
showing that the increase in acceleration would decrease as $r^{-3}$, which is indeed what the above geodesic equation approaches in the limit $r \to \infty$.
The redshift effects cause our solar sail to 'speed up' much faster than $r^{-3}$ when we get closer to the star (as the sail now get's an additional boost from what it sees as blueshifted photons closer to the star).
A: Kerzerashvili and Vázquez-Poritz has a nice paper about solar sails in general relativity. They derive the orbital equation $$\left ( \frac{dr}{d\phi}\right )^2 = \left ( \frac{E^2}{c^2} - c^2 + \frac{2G\tilde{M}}{r} - \frac{L^2}{r^2}f\right ) \frac{r^4}{L^2}$$
where $f=1-\frac{2GM}{c^2r}$ and $\tilde{M}=M-\kappa/G$ is a renormalized mass adjusted for the $1/r^2$ decay of both gravity and light pressure (note that they do not include the gravitational redshift), and $\kappa$ is the factor due to light reflection/absorption. $$\kappa = [(2\eta-1)\cos(\alpha+\psi)\cos(\psi)-(1-\eta)\sin(\alpha)]\lambda \cos(\alpha+\psi)$$ where $\lambda=L_\odot/2\pi c \sigma$, $\alpha=\sin^{-1}(v/c)$ the aberration due to speed, and $\psi$ the solar sail tilt angle. Phew. 
For a statite $\alpha=\psi=0$, $\kappa=(2\eta-1)\lambda$, $dr/d\phi=0$, $L=0$ and the stability becomes $$r = \frac{2G\tilde{M}}{c^2 - E^2/c^2}.$$ So depending on the energy you get more remote orbits, with $E=0$ implying the event horizon "orbit". 
For circular orbits they derive a period $$T^2=\frac{4\pi^2 r^3}{G\tilde{M}}\left [ 1+ \kappa \frac{c^2r-4GM}{(c^2r - 2GM)^2}\right ].$$ which behaves just as Kepler's law if $\kappa=0$. The leading correction is $$T^2\approx  \frac{4\pi^2 r^3}{G\tilde{M}}\left [ 1+ \kappa/c^2r\right ]$$ which implies that close to the star orbits will be slower than in Newtonian gravity - the effect is as if gravity was weaker ($\tilde{M}<M$) and there is a tiny bit of extra distance to go ($\kappa/c^2r$).
There is another factor due to frame dragging affecting the period slightly, but it can go either way depending on whether the orbit is prograde or retrograde. They even include a cosmological constant, which adds a tiny $(c^2\Lambda/2G\tilde{M})r^3$ term inside the bracket. 
So I guess one can say that there is a repulsion-like effect increasing for small $r$.
Now, if we consider the gravitational redshift when going from $r_1$ to $r_2$ it will be $$\lambda_2 = \lambda_1\sqrt{\frac{r_1(r_2-R_s)}{r_2(r_1-R_s)}}$$ or, for the ratio of received energy, $\sqrt{\frac{r_1}{r_1-R_s}}\sqrt{1-\frac{R_s}{r_2}}=(\sqrt{\frac{1}{r_1}}+\ldots )(1-\frac{R_s}{2r_2}+\ldots) \approx \frac{1-\frac{R_s}{2r_1}}{\sqrt{r_1}} $ where the series expansion assumes $r_{1,2}$ are large compared to $R_s$ (the next term for $r_1$ is $(R_s/2)r_1^{-3/2}$). So we get a reduction of $L_\odot$ that scales like $L(r_2)\approx (1-R_s/r_2)L_\odot/\sqrt{r_1}.$ Now, in the equations above this just implies $\kappa$ gets reduced and hence $\tilde{M}$ approaches $M$, but the stability equation just shifts a little bit for a given $E$ - there is still an equilibrium.
