Random Noise Voltage
I hate to be a cold blanket, but you're pretty much out of luck for any normal definition of "useful power".

You have a signal which produces ~ 2v pk-pk with a source impedance of 3.5 k. There is a well-known relationship that states/shows that you'll get maximum power into a load when the load impedance is equal to the source impedance. So in this case, the peak power you might expect would be something like







simulate this circuit &ndash; Schematic created using CircuitLab

and it should be pretty obvious that the maximum power will be $$P = \frac{V^2}{R} = \frac{0.5^2}{3500} = 0.07\text{ mW} $$ or 70 microwatts. If you don't get the numbers, the voltage on the load will be 0.5 because the source impedance cuts it in half.

And that's the peak. RMS is one quarter that (since the rms voltage is half the peak), or about 17 microwatts. And any real collection device will have efficiencies less than one. 

Are you sure that 10 microwatts or so is worth your time and trouble?

And all of this is before you get into the problems of things like diode drops which will make your life really miserable. So 10 microwatts is actually wildly optimistic.