= Solution
The <effective mass> follows from the curvature of the <electronic band structure>. Differentiating twice gives
$$
\frac{dE}{dk}=\frac{\hbar^2}{m_ea}\sin ka,\qquad
\frac{d^2E}{dk^2}=\frac{\hbar^2}{m_e}\cos ka,
\qquad m_e^*(k)=\frac{m_e}{\cos ka}.
$$
At either first <Brillouin zone> boundary $k=\pm\pi/a$,
$$
\boxed{m_e^*=-m_e.}
$$
The boundary is a band maximum, with negative curvature and zero <group velocity>. In the semiclassical single-band description, an applied <force> changes $\hbar k$; the resulting <acceleration> is opposite the <force> near that maximum. This is a band-response coefficient, not a negative microscopic <electron> rest mass. A nearly filled band is often described more conveniently through positively charged <electron holes> near its maximum.
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