= Solution
Measure single-particle <energy> from the lowest state and write $\varepsilon_k=c k^\alpha$, with $c>0$. In a homogeneous system, the number of <wavevector> states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\varepsilon$ gives the <density of states> per unit spatial <volume (thermodynamics)>,
$$
D(\varepsilon)=\frac{S_{d-1}}{(2\pi)^d\alpha}\,c^{-d/\alpha}\varepsilon^{d/\alpha-1},
$$
where $S_{d-1}$ is the <area> of the <unit sphere> in $d$ dimensions; an additional internal <state degeneracy> would multiply this expression by a constant.
The <Bose-Einstein distribution> has $\mu\leq0$. At saturation, $\mu\to0^-$, the maximum excited-state <number density> is
$$
n_{\mathrm{ex}}^{\max}(T)=\int_0^\infty\frac{D(\varepsilon)}{e^{\beta\varepsilon}-1}\,d\varepsilon,
\qquad\beta=\frac1{k_BT}.
$$
Only the lower endpoint can cause divergence. There the integrand is proportional to
$$
\frac{\varepsilon^{d/\alpha-1}}{\beta\varepsilon}
\propto\varepsilon^{d/\alpha-2}.
$$
The <improper integral> converges precisely when its power exceeds $-1$. Thus
$$
\boxed{d>\alpha\quad\text{is the condition for }T_c>0.}
$$
When $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the <excited states> can accommodate any fixed finite density at every positive <temperature>, so there is no bulk <Bose-Einstein condensation> at nonzero <temperature>.
For completeness, putting $s=d/\alpha>1$ and using the <Gamma function> and <Riemann zeta function> gives
$$
n_{\mathrm{ex}}^{\max}(T)=\frac{S_{d-1}}{(2\pi)^d\alpha}c^{-s}\Gamma(s)\zeta(s)(k_BT)^s.
$$
At fixed density $n$, the right-hand side reaches $n$ at $T_c$; below it the excess particles occupy the lowest state, with <condensate fraction> $1-(T/T_c)^s$. This is <Bose condensation with power-law dispersion> in the homogeneous <thermodynamic limit>. A finite vessel or a trapping potential changes the low-energy state count and must not be silently substituted for that setting.
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