Without more details, when a signal is properly recorded at $F_n$ Hz, one is entitled to expect that all information below the half of it, namely $F_n/2$ Hz, can be recovered, somehow. This is an application of the classical Shannon Sampling Theorem. Twice the maximum frequency is sufficient to sample a signal... in theory. The question says: max frequency that human voice can be captured; don't expect this max frequency (and others) be capture with full accuracy, as suggested in comments by David.
In some cases, this can be extended, as long as the bandwidth can be shrunk. If the actual bandwidth is smaller, like $F_n/4$-$F_n/2$, one might expect to sample it at $2\times (F_n/2-F_n/4) = F_n/2$ without loss, yet I don't think it is the expected answer, cf. Voice Frequency :
The bandwidth allocated for a single voice-frequency transmission
channel is usually 4 kHz, including guard bands, allowing a sampling
rate of 8 kHz to be used as the basis of the pulse code modulation
system used for the digital PSTN. Per the Nyquist–Shannon sampling
theorem, the sampling frequency (8 kHz) must be at least twice the
highest component of the voice frequency via appropriate filtering
prior to sampling at discrete times (4 kHz) for effective
reconstruction of the voice signal.