= Solution
For brevity write $\gamma=mKZ/\hbar^2$. With the physical <indicial root> $c=\ell+1$, substitute $f=\sum_{n\geq0}a_nr^{n+\ell+1}$ into the equation from part (i). The coefficient of $r^{n+\ell}$ is
$$
\bigl[(n+\ell+2)(n+\ell+1)-\ell(\ell+1)\bigr]a_{n+1}+
\bigl[2\gamma-2\alpha(n+\ell+1)\bigr]a_n=0.
$$
The <indicial equation> has already canceled the still lower power $r^{\ell-1}$. Rearranging gives the <recurrence relation>
$$
\boxed{a_{n+1}=2a_n\frac{\alpha(n+\ell+1)-mKZ/\hbar^2}{(n+\ell+2)(n+\ell+1)-\ell(\ell+1)}.}
$$
Its denominator is $(n+1)(n+2\ell+2)>0$ for every $n\geq0$.
To impose <wavefunction normalization> at infinity, first suppose that the series never terminates. Then
$$
\frac{a_{n+1}}{a_n}\sim\frac{2\alpha}{n},\qquad n\to\infty.
$$
This is the exponential-series growth associated with $e^{2\alpha r}$, which would overwhelm the factor $e^{-\alpha r}$ in $\psi$. A direct lower-bound argument makes the conclusion precise. All sufficiently late recurrence ratios are real and positive, so after a common <global phase> choice the tail coefficients have one sign. For any $0<\eta<\alpha$, there is an index $N$ such that
$$
\frac{|a_{n+1}|}{|a_n|}\geq\frac{2\alpha-\eta}{n+1},\qquad n\geq N.
$$
Induction then bounds the tail below by a positive multiple of
$$
r^{\ell+1}\sum_{n\geq N}\frac{[(2\alpha-\eta)r]^n}{n!}.
$$
The finitely many earlier terms form a <polynomial> and cannot cancel this exponential growth. Thus $|\psi|=r^{-1}e^{-\alpha r}|f|$ grows at least as a positive multiple of $r^\ell e^{(\alpha-\eta)r}$ for large $r$, contradicting <wavefunction normalization>. This proves that the regular series must terminate.
Since the denominator never vanishes, termination requires the numerator to vanish at some nonnegative integer $n$:
$$
\boxed{\alpha=\frac{mKZ}{(n+\ell+1)\hbar^2}.}
$$
Then $a_{n+1}=0$ and every later coefficient is zero. Consequently \b[$f$ is a polynomial of degree $n+\ell+1$], with a factor $r^{\ell+1}$; equivalently,
$$
f(r)=r^{\ell+1}P_n(r),\qquad \psi(r)=r^\ell P_n(r)e^{-\alpha r},
$$
where $P_n$ has degree $n$. This is <series-termination quantization of the Coulomb radial equation>. If $N_{\rm p}=n+\ell+1$ denotes the <principal quantum number>, the associated <energy eigenvalues> are
$$
\boxed{E_{N_{\rm p}}=-\frac{mK^2Z^2}{2\hbar^2N_{\rm p}^2},\qquad N_{\rm p}=1,2,\ldots.}
$$
Here $n$ counts the degree of the remaining radial polynomial; it is not the principal <quantum number> itself.
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