= Solution
Expand a single-electron <wavefunction> as $|\psi\rangle=\sum_n\psi_n|n\rangle$. Project the <Schrödinger equation> onto $\langle n|$ to obtain
$$
E\psi_n=\epsilon_0\psi_n-t\psi_{n-1}-t\psi_{n+1}.
$$
Translation <symmetry> suggests the <Bloch state> $\psi_n=N^{-1/2}e^{ikna}$ on a periodic chain of $N$ sites. Substitution gives
$$
\boxed{E(k)=\epsilon_0-t(e^{ika}+e^{-ika})=\epsilon_0-2t\cos(ka).}
$$
Take the conventional positive hopping $t>0$. The first <Brillouin zone> is $-\pi/a\leq k<\pi/a$; the band has minimum $\epsilon_0-2t$ at $k=0$ and maximum $\epsilon_0+2t$ at the zone edges. Its width is $4t$. A negative hopping interchanges the extrema but does not change the <density of states> after replacing $t$ by $|t|$.
There are $N$ allowed <wavevectors> and two <spin> states per <wavevector>. For the total spin-summed <density of states> $D(E)$, each interior band <energy> has two roots, $\pm k$. Using the spacing $2\pi/(Na)$ gives
$$
D(E)=\frac{2Na}{2\pi}\sum_{k:E(k)=E}\frac1{|dE/dk|}
=\frac{2N}{\pi\sqrt{4t^2-(E-\epsilon_0)^2}}.
$$
Thus the <density of states> per atom is
$$
\boxed{g(E)=\begin{cases}\dfrac{2}{\pi\sqrt{4t^2-(E-\epsilon_0)^2}},&|E-\epsilon_0|<2t,\\0,&|E-\epsilon_0|>2t.\end{cases}}
$$
It is symmetric about $\epsilon_0$ and has integrable inverse-square-root <Van Hove singularities> at both band edges. Its <integral> over the band is two states per atom. The plot labels the dispersion extrema and both singular edges.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018-tight-binding-band-and-density.png]
{title=One-dimensional <tight binding> dispersion and spin-summed <density of states>}
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