Understanding FFT bin resolution For a signal sampled at $f_s$, the frequency resolution (or "bin width") for an $N$ point FFT is $f_s/N$. Does this mean that the $k$th bin will contain energy from sinusoids within $0.5f_s/N$ on either side of the center frequency of that bin (ignoring spectral leakage for now)? Is this energy averaged? Is it the same as taking a $2N$ point FFT and then averaging every two bins?
The energy bin "width" of an FFT is the width of the transform of the window used (a rectangular window by default).  

The transform of a rectangle the width of the FFT is a Sinc function (sin(x)/x) with a main lobe width equal to twice the DFT/FFT bin spacing, plus smaller ripples extending the full width of the FFT result (and actually circling around forever making it technically a Dirichlet function).  Note that the frequency response of each bin filter overlaps with others except at exact bin centers (representing exactly periodic-in-aperture pure sinusoids).

For the default rectangular window, only about 87% of a bin's passband is within 0.5 of a bin width from the bin center frequency, which is why ignoring so-called leakage might not be a good idea.  A longer FFT has a narrower main lobe bin response, with the ripples ("leakage") dying out faster.  But 2 half width Sinc functions do not sum to a single Sinc, so averaging a longer FFT will not produce identical results (except for sinusoids that are exactly periodic in both FFT lengths.)

A window other than rectangular (von Hann, etc.) will have a main lobe wider that 2 bin spacings, but, in return, far smaller ripples away from the main lobe.

The DFT/FFT resolution depends on whether you are trying to resolve closely spaced peaks as clearly separate peaks (requiring around double or more the bin spacing), or estimating the frequency of a single isolated peak with a relatively high local S/N ratio (sometimes in low noise allowing a fraction of a bin spacing in resolution after high quality interpolation).