Blinking flashlight infalling into the black hole, as frequency → ∞

The coordinate acceleration by proper time for a radially infalling observer is simply the Newtonian expression

$$ \ddot{r} = \frac{{\rm d}^2 r}{{\rm d} \tau^2} = -\frac{G M}{r^2}$$

which integrates to

$$ \dot{r} = \frac{{\rm d} r}{{\rm d} \tau} = -\sqrt{\frac{2 G M (r_1-r)}{r_1 \ r}} $$

so if you start at rest at $r_1$ and freely fall to $r_2$ the elapsed proper time is

$$ \tau = \int_{r_1}^{r_2} \frac{1}{\sqrt{2 G M \left(\frac{1}{r}-\frac{1}{r_2}\right)}} \, \text{d}r$$

The time dilation relative to the coordinate bookkeeper far away from the black hole is

$$ \dot{t} = \frac{{\rm d} t}{{\rm d} \tau} =\sqrt{\frac{c^2 \ r-c^2 \ r_s +r \ \dot{r}^2}{c \ (r- r_s ) \ (1-r_s/r)}} $$

For example let's plug in some numbers; if you fall from the photon sphere at $r_1=1.5 \ r_s$ you reach the horizon at $r_2=r_s=2G M/c^2$ when your proper time is $\tau=3.993468$ so if you set the intervall of your flashlight to give $100$ pulses per $1GM/c^3$ proper time, the last received pulse is that given at $\tau=3.99 \ G M/c^3$ when the flash light is at $r=2.001999 \ G M/c^2$, which is at coordinate time $t=18.43309 \ G M/c^3$.

If you want to know when this last pulse of light is observed you have to add the light travel time; the shapirodelayed coordinate velocity of a radially outgoing light ray is

$$ r' = \frac{{\rm d} r}{{\rm d} t} = c-\frac{2 G M}{r c} $$

so the light travel coordinate time from where that signal was emmitted to the observer at $r_1=3 G M/c^2$ is $t_{\rm \ light} = 13.42798 \ G M/c^3$, so the last signal which is emmitted outside of the horizon will be received at $t_{\rm obs}=t+t_{\rm \ light}=31.86107 \ G M/c^3$, or in terms of the proper time of the stationary shell observer at $r=r_1$ that $399 \rm th$ and last signal would be received at ${_{T}}_{\rm obs}=t_{\rm obs} \sqrt{1-r_s/r_1}=18.395 \ G M/c^3$.

If you set the frequency of your flashlight to the limit of infinitely many pulses per unit proper time, the last received signal will also be seen only after an infinite coordinate- or shell-time.