The Schwarzchild geometry is defined as
$$ds^2=-\left(1-\frac{2GM}{r} \right)dt^2+\left(1-\frac{2GM}{r} \right)^{-1}dr^2+r^2(d\theta^2+\sin^2(\theta) d\phi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=\infty$, we get
$$d\tau^2=-ds^2=\left(1-\frac{2GM}{\infty} \right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $\theta=\pi/2$) a distance $r=r_0$ away from the black hole, we get
$$d\tau^2=\left(1-\frac{2GM}{r_0} \right)dt^2-{r_0}^2d\phi^2$$
For a circular orbit, it can be shown that $r_0^2 d\phi^2=\frac{GM}{r_0}dt^2$ and hence
$$d\tau^2=\left(1-\frac{3GM}{r_0} \right)dt^2$$
Thus $d\tau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?