= Solution
Write
$$
f(r)=1-\frac{2M}{r}.
$$
For a static observer, $d\tau=\sqrt{f(r)}\,dt$. Conservation of the photon's Killing energy therefore gives the <gravitational redshift between static Schwarzschild observers>
$$
\frac{\lambda_{\rm obs}}{\lambda}
=\sqrt{\frac{f(r)}{f(r_{\rm st})}}.
$$
For the stated distant observer, $r\gg r_{\rm st}$ and $f(r)\simeq1$, so
$$
\boxed{
\lambda_{\rm obs}
\simeq\frac{\lambda}{\sqrt{1-2M/r_{\rm st}}}
}.
$$
As $r_{\rm st}\to2M^+$, the observed wavelength tends to infinity: light emitted from a static source arbitrarily near the <Schwarzschild event horizon> is infinitely redshifted at infinity.
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