What's the theoretical bandwidth?
This question seems to highlight some of the perils of translating concepts between the continuous domain and a discrete one.  My understanding is that bandwidth is defined in the continuous domain as the difference between the upper and lower band limits.  Since it is the interval length, whether the interval is open or closed doesn't matter.  Open or closed means if the endpoints are considered part of the interval.  Since we are always saying "up to the Nyquist", considering the endpoints to be open seems logical.

The extra twist in this is your stipulation that the DC component is zero.  This would seem to narrow the bandwidth at the lower end, but to what?  In the continuous case, it would be an infinitesimal amount.

Can this properly be translated to the discrete case you describe?  I don't think so.  There may be a common convention for doing so that I am unaware of.  Perhaps your approach, but then you should strike off the gap at both ends, since you have excluded  DC.

Upon some reflection, I am going to disagree with MBaz again, and say that $f_s/2$ is the theoretical accurate answer based on the continuous definition of bandwidth.

I will agree that these are purely theoretical considerations with negligible practical value, but then again, so are a lot of exam questions.

You should probably strike the word "radio", it stopped RB-J from giving a good answer.  Probably a better answer than what I just gave.  



$$ H_k(\omega) = \sum_{n=0}^{N-1}W_N^{-kn}e^{-jn\omega}  $$

$$ W_N^{-kn} = \left( e^{-j \frac{2\pi}{N}} \right) ^{-kn} = e^{j kn \frac{2\pi}{N}} $$

$$
\begin{aligned}
H_k(\omega) &amp;= \sum_{n=0}^{N-1}e^{j kn \frac{2\pi}{N}}e^{-jn\omega} \\
 &amp;= \sum_{n=0}^{N-1} e^{-jn ( -k \frac{2\pi}{N} + \omega) } \\
 &amp;= \sum_{n=0}^{N-1} \left( e^{-j ( -k \frac{2\pi}{N} + \omega )} \right)^n \\
 &amp;= \frac{1 - \left( e^{-j ( -k \frac{2\pi}{N} + \omega )} \right)^N }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\
 &amp;= \frac{1 - e^{-j ( -k 2\pi + \omega N )}  }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\
 &amp;= \frac{1 - e^{-j \omega N }  }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\
 &amp;= \frac{e^{-j \frac{\omega N}{2} }}{e^{-j ( \frac{-k\pi}{N} + \frac{\omega}{2} )}} 
 \cdot \frac{ e^{j \frac{\omega N}{2} } - e^{-j \frac{\omega N}{2} }  }{ e^{j ( \frac{-k\pi}{N} + \frac{\omega}{2} )} - e^{-j ( \frac{-k\pi}{N} + \frac{\omega}{2} )} } \\
 &amp;= e^{j \left( \frac{-k\pi}{N} - \frac{\omega (N-1)}{2} \right) } 
 \cdot \frac{ \sin \left( \frac{\omega N}{2} \right) }{ \sin \left(  \frac{-k\pi}{N} + \frac{\omega}{2}  \right) } \\
\end{aligned}
$$

For similar math, see my blog article DFT Bin Value Formulas for Pure Complex Tones.  The section called "As N Gets Large" covers the relation between "alias" sinc, aka Dirichlet Kernel, and the normalized sinc function.