= Solution
Let $u=rF$. Multiplication of the radial equation by $-2mr/\hbar^2$ gives $u''+[A/r-\lambda(\lambda+1)/r^2-\alpha^2]u=0$. With $u=e^{-\alpha r}f$ the constant terms cancel, leaving
$$
\boxed{f''-2\alpha f'-\lambda(\lambda+1)f/r^2+Af/r=0.}
$$
The <indicial equation> is $c(c-1)=\lambda(\lambda+1)$, with roots $c=\lambda+1$ and $c=-\lambda$. The usual physical <Coulomb radial origin condition> selects $c=\lambda+1$: it imposes the regular reduced radial boundary condition $u(0)=0$. Normalizability alone at the origin is insufficient to make this selection when an irregular branch is locally <square-integrable>, notably at $\lambda=0$.
Write $f=r^{\lambda+1}\sum_{j\geq0}a_jr^j$. Coefficient comparison gives
$$
a_{j+1}=\frac{2\alpha(j+\lambda+1)-A}{(j+1)(j+2\lambda+2)}a_j.
$$
A nonterminating regular series has the exponentially growing asymptotic branch, suggested by $a_{j+1}/a_j\sim2\alpha/j$: it supplies an $e^{2\alpha r}$ component, which the prefactor $e^{-\alpha r}$ cannot normalize. Equivalently the regular confluent-hypergeometric solution grows unless its first parameter is a nonpositive integer. Thus a bound state requires the numerator to vanish at some $j=k\geq0$, giving
$$
\boxed{\alpha=\frac{A}{2(k+\lambda+1)},\qquad E=-\frac{\hbar^2A^2}{8m(k+\lambda+1)^2}.}
$$
This is <series-termination quantization of the Coulomb radial equation>. It requires $A>0$ for the assumed $\alpha>0$; at $A=0$ there are no such negative-<energy> Coulomb bound states.
For $k=0$, $F=Nr^\lambda e^{-\alpha r}$. The angular integral cancels when computing $\langle1/r\rangle$, while the radial volume element contributes $r^2dr$. Consequently
$$
\left\langle\frac1r\right\rangle=\frac{\int_0^\infty r^{2\lambda+1}e^{-2\alpha r}\,dr}{\int_0^\infty r^{2\lambda+2}e^{-2\alpha r}\,dr}=\frac{\alpha}{\lambda+1}.
$$
The ratio follows by one <integration by parts> and remains valid for real $\lambda\geq0$; no integer-only factorial identity is needed. Since now $\alpha=A/[2(\lambda+1)]$,
$$
\boxed{\langle V\rangle=-\frac{\hbar^2A\alpha}{2m(\lambda+1)}=-\frac{\hbar^2A^2}{4m(\lambda+1)^2}=2E.}
$$
For a classical particle with that negative total <energy>, nonnegative <kinetic energy> requires $E-V=\hbar^2(A/r-\alpha^2)/(2m)\geq0$, hence $\boxed{r\leq A/\alpha^2}$. This is a necessary radial bound; extra angular momentum can impose a stronger restriction.
At $k=\lambda=0$, $\alpha=A/2$ and the classical boundary is $r=2/\alpha$. The normalized radial <probability> density is $4\alpha^3r^2e^{-2\alpha r}$. With $z=2\alpha r$, the exterior quantum <probability> is
$$
\boxed{\int_{2/\alpha}^\infty4\alpha^3r^2e^{-2\alpha r}\,dr=\frac12\int_4^\infty z^2e^{-z}\,dz=13e^{-4}.}
$$
The final integration gives $(4^2+2\cdot4+2)e^{-4}/2$. This is the nonzero quantum tail beyond the classical turning region.
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