The particle horizon is given by \begin{equation} \chi_p(\eta) = \int_{t_i}^{t}\frac{dt}{a} = \int_{a_i}^{a}\frac{da}{a\dot{a}} = \int_{\ln a_i}^{\ln a} (aH)^{-1}\ d\ln a . \end{equation}
For a universe dominated by a fluid with constant equation of state $w = P/\rho$ we have, \begin{equation} (aH)^{-1} = H_0^{-1}a^{\frac{1}{2}(1 + 3w)} . \end{equation}
We can calculate the the particle horizon explicitly for this case, \begin{equation} \chi_p(a) = \frac{2H_0^{-1}}{(1 + 3w)} \left[ a^{\frac{1}{2}(1 + 3w)} - a_i^{\frac{1}{2}(1 + 3w)} \right] \equiv \eta - \eta_i \end{equation}
For familiar matter sources ($1 + 3w > 0$) the largest contribution comes from later times, i.e. \begin{equation} \eta_i = \frac{2H_0^{-1}}{(1 + 3w)}a_i^{\frac{1}{2}(1 + 3w)} \longrightarrow 0 \quad \text{for}\quad a_i\to 0,\ w > -\frac{1}{3} . \end{equation}
Therefore, the comoving horizon is finite \begin{equation} \chi_p(t) = \frac{2H_0^{-1}}{(1 + 3w)}a(t)^{\frac{1}{2}(1 + 3w)} = \frac{2}{(1 + 3w)}(aH)^{-1} . \end{equation}
Using this as motivation, Baumann and Dodelson suggest that the natural solution to the horizon problem is to conjecture a phase of decreasing Hubble radius in the early universe,
$$ \frac{d}{dt}(aH)^{-1} < 0. $$
This is cited as the solution because the Hubble sphere is growing in standard Big Bang cosmology.
Now, I don't understand why it is a natural or easy solution to consider a period of time where the Hubble radius is decreasing. Is there some key insight that I am missing? Any help would be appreciated.