Does time speed up or slow down near a black hole? 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?
 A: You are mixing the two times up. Proper time $\tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
A: To expand on Javier's answer, the symbol $\tau$ represents proper time, i.e. time as measured in a particular reference frame.  You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer.  Equating these two different proper times is incorrect, which is why you are getting the wrong answer.  The quantity that does not change between frames is $t$, which is the coordinate time.  In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $\tau_\infty$ for proper time in the frame of the stationary observer and $\tau_{orbit}$ for proper time in the frame of the orbiting observer.  Then as you correctly showed,
\begin{align}
d\tau_\infty^2 &= dt^2 \\
d\tau_{orbit}^2 &= \left( 1-\frac{3GM}{r_0} \right) dt^2
\end{align}
It follows that
$$d\tau_{orbit}^2 = \left( 1-\frac{3GM}{r_0} \right) d\tau_\infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
A: As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$\tau$. His final formula is right but its interpretation is upside
down.
