= Solution
Define <outgoing Eddington-Finkelstein coordinates> by $u=t-r_*$. The radial <Schwarzschild metric> becomes
$$
ds^2=-f\,du^2-2\,du\,dr.
$$
Outgoing radio rays have constant $u$. Two nearby wave crests emitted at Alice's <proper times> $\tau$ and $\tau+d\tau$ have separation $du=\dot u_A\,d\tau$. At Bob's fixed radius $r_B$, $dt_B=du$ and $d\tau_B=\sqrt{f_B}\,du$, where $f_B=1-2M/r_B$. Thus the exact local <redshift> is
$$
\boxed{\frac{\lambda_B}{\lambda_A}=\frac{d\tau_B}{d\tau_A}=\sqrt{f_B}\,\frac{du_A}{d\tau_A}.}
$$
Here the <wavelength> ratio equals the <proper time> period ratio in <geometric optics>. Equivalently, the phase gradient $k_a=-\omega_\infty(du)_a$ gives Alice's measured <frequency> $\omega_A=-k_aU_A^a=\omega_\infty\dot u_A$ and Bob's $\omega_B=\omega_\infty/\sqrt{f_B}$.
\b[In the asymptotically distant-observer limit $f_B\to1$, this becomes $\lambda_B/\lambda_A=du_A/d\tau_A$.] “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.
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