= Solution
Put $A=n^2-1/4$. Substitute $w=e^{sz}\sum_{k\geq0}c_kz^{\rho-k}$ into $w''+(1-A/z^2)w=0$. The leading coefficient gives $s^2+1=0$, and the next gives $2s\rho c_0=0$. Thus $s=\pm i$, $\rho=0$. The coefficient of $e^{sz}z^{-k-2}$ then gives
$$
\boxed{c_{k+1}=\frac{k(k+1)-A}{2s(k+1)}c_k,\qquad c_0\ne0.}
$$
In particular $c_1=sAc_0/2$ and $c_2=A(2-A)c_0/8$. The two <Bessel inverse-power expansions at infinity> start as
$$
\boxed{w_\pm\sim C_\pm e^{\pm iz}\left(1\pm\frac{iA}{2z}+\frac{A(2-A)}{8z^2}+\cdots\right).}
$$
Their <independent> leading oscillations give two <independent> asymptotic solutions, represented by suitable multiples of $\sqrt zH_n^{(1)}(z)$ and $\sqrt zH_n^{(2)}(z)$ on the positive-real large-$z$ sector.
For the <Liouville–Green approximation>, $q=1-A/z^2$ gives
$$
q^{-1/4}=1+\frac A{4z^2}+O(z^{-4}),\qquad
\int^z\sqrt q\,d\zeta=z+\frac A{2z}+O(z^{-3}),
$$
where the integration constant is absorbed in the amplitude. Hence its two leading wave forms have first two terms
$$
\boxed{q^{-1/4}e^{\pm i\int^z\sqrt q}
=e^{\pm iz}\left(1\pm\frac{iA}{2z}+O(z^{-2})\right).}
$$
Expanding one order further gives the same $A(2-A)/(8z^2)$ as the formal recurrence.
The infinite sums must be interpreted as <asymptotic expansions>, not literal convergent inverse-power series. For integer $n$, $A$ never equals $k(k+1)$, so the recurrence never truncates, and $|c_{k+1}/c_k|\sim k/2$. Thus the inverse-power series diverges at every finite nonzero $z$. Exact Hankel solutions have these asymptotic series; the printed “solutions of the form” uses that customary asymptotic meaning.
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