If you're watching a live stream and n number of minutes behind the live feed, how much time would it take while watching at s speed to catch up? Sometimes I'll tune into YouTube livestreams a few minutes late, but I want to watch the whole thing and catch up with the live feed, so I'll start from the beginning at 2x speed or 1x speed. I'd like to find a way to calculate exactly how much time it will take for me to reach the live feed.
So if I start watching 20 minutes behind at 2x speed, after 20 minutes of watching, I'll be 10 minutes behind, after 10 minutes, I'll be 5, etc.
I would guess this would look something like this
$ total = \frac{20}{2} + \frac{10}{2} + \frac{5}{2} + ... $
How would you create a general equation for this? Could you use summation, product, or something like a Taylor series? I recall an old VSauce Video about SuperTasks which feels relevant here as well.

First, analogize to chasing a piece of driftwood that is in a stream, whose current is going $1$ mile per hour.  Assume that you are running parallel to the driftwood, along the bank of the stream, at $x$ miles per hour.  Also assume that you start out $k$ miles behind the driftwood.
Since you are gaining $(x-1)$ miles, for ever hour that you run, and since you start out $k$ miles behind the driftwood, it will take you $~\dfrac{k}{x-1}~$ hours to catch the driftwood.
The analogy is apt.  Assume that you start out $k$ minutes behind the show.  Also assume that you watch the show at $r$ times the normal speed.  For example, $r=2$ implies that you are watching the show at twice the normal speed.
This means that every minute that the show progresses by $1$ minute, you progress by $r$ minutes.  This implies that you are catching up to the metaphoric driftwood, at a rate of $(r-1)$ minutes per minute.
Note, that here, the distance to catch up is measured in minutes, rather than (for example) miles.
So, you are catching up by $(r-1)$ minutes, for every minute that you watch, and the driftwood starts out $k$ minutes ahead.
So, it will take you $~\dfrac{k}{r-1}~$ minutes to catch up.
