Why do smaller wavelengths image particles more clearly when the particle size is already big? Yesterday I scanned an ultrasound phantom that had cylindrical inclusions (1.5mm diameter).  When I boosted the ultrasound frequency from 3MHz to 4MHz, these inclusions became much more clear.  I understand that the resolution of the system is limited by the wavelength of the sound wave.  However, I calculated the wavelength to be about 1/3 the size of the inclusion (0.5mm and 0.4mm).  Why does wavelength still improve detectability and clarity even though the wavelengths are less than the inclusion size?
 A: You wrote, 

Why does wavelength still improve detectability and clarity even though the wavelengths are less than the inclusion size?

The smallest feature resolvable by a typical ultrasound system is 1/2 of the wavelength of the sound used in the system, in your case the wavelengths of 3 MHz and 4 MHz waves in your medium.  Assuming the sound speed is the same for both frequencies, the smallest features resolvable by a 4 MHz system should be 3/4 of the size of features resolvable by a 3 MHz system.
Note that an edge or a corner of an inclusion, to be recognizable as such, has crucial features that are much smaller than the whole inclusion. So, the wavelength required to image sharply a corner or edge is much shorter than the wavelength needed to image a blob vaguely the same shape as the inclusion.
If the inclusion is smaller than the wavelength of the sound, it will not be resolvable at all.  That is, 1/2 the sound wavelength must be smaller than the inclusion or the inclusion will not be resolved at all, even as a fuzzy blob.  If I understand your question correctly, you can resolve the inclusions at 3 MHz, but their shapes look more recognizable at 4 MHz.
Consider the resolution that would be needed to get a good sense of the shape of an inclusion. The idea of "resolution" is closely analogous to pixel size.  If a photo of the inclusion has pixels as big as the inclusion, the photo will give you no information about the shape of the inclusion, just about its location.  With pixels about 1/4 of the size of the inclusion you might start to get some sense of the shape.  At 1/10 of the size of the inclusion you could get a much better sense of the shape.   
Another analogy:  Imagine the character "8" represented with a resolution of 1 x 2 pixels.  It would just be a gray rectangle.  With a resolution of 2 x 4 pixels, it again would just be a gray rectangle.  With a resolution of 3 x 6 pixels, it would suddenly look like a very fuzzy version of the figure "8".  With a resolution of 4 x 8 pixels it would be easily recognizable as a figure "8". (Single-digit numbers displayed on a 4 x 8 pixel display are easily distinguishable.)
This really doesn't have anything to do with harmonics; the resolution is directly related to the wavelength of the acoustic signal used to form the image. Harmonics of the signal are not included in the ultrasound imaging process.
A: A ratio of 1 to 3 is well within the regime where the wavelength is "similar" to the object size, and diffraction/scattering effects are massively important.
Without even getting into the details of diffraction patterns (which make the difference between these frequencies even more pronounced than my simplified explanation suggests), if we suppose that the resolution of the image produced will make the smallest useful pixel size about a wavelength wide, then at one frequency you have 3 pixel wide cylinders, and at the other you have 4 pixel wide cylinders.  That is a huge difference in the amount of visible information about the cylinders.
For a more detailed discussion of ultrasound resolution, see for example section 1.2 of Practical Clinical Ultrasonic Diagnosis By Liwu Lin (it can be found on google books).  It explains that longitudinal resolution usually approaches the theoretical limit of one wavelength, and that transverse resolution is generally much worse.  So you're lucky to even see the inclusions at these wavelengths, let alone see a clear shape!
