 Solution

Define outgoing Eddington-Finkelstein coordinates by $u=t-r_{\ast }$. The radial Schwarzschild metric becomesOutgoing 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 $r_B$, $d{t}_B=du$ and $d{\tau }_B=\sqrt{f_B}du$, where $f_B=1-2M/r_B$. Thus the exact local redshift isHere 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 ${\omega }_A=-k_a{U}_A^a={\omega }_{\infty }{\dot{u}}_A$ and Bob's ${\omega }_B={\omega }_{\infty }/\sqrt{f_B}$.

In the asymptotically distant-observer limit $f_B\to 1$, this becomes ${\lambda }_B/{\lambda }_A=d{u}_A/d{\tau }_A$. “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.