How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing? I am just getting into the field of cosmology and was wondering what it means practically to say that the Hubble Constant is decreasing while the expansion rate of the Universe is accelerating. I am not sure how this phenomenon is possible. How do we observe this? 

It is my understanding that the Hubble Constant is the rate at which the Universe is expanding, and it is decreasing with time. (Although it is asymptotic)... Is this conception flawed?
The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.
The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.

In the future, the cosmological constant, $\Lambda$ could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.

For example, suppose $H(t) = (da(t)/dt)/a(t)$, where $a(t)$ is the distance between two arbitrary points in the universe at epoch $t$. Now let's have an expansion rate $da/dt \propto t$ (ie accelerating). But in this case $a(t) \propto t^2$ and $H(t) \propto t ^{-1}$ (ie decreasing with time).

Some extra details: 

The solution to the Friedmann equation in a flat universe is
$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$
where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.

As the universe expands, $\rho$ of course decreases as $a(t)^{-3}$, but $\Lambda$ remains constant. Thus the first term on the RHS becomes unimportant if indeed $\Lambda$ is a cosmological constant.

Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.