TITLE: How long to catch up to a stream started 1 hour ago at 1.5x speed?
QUESTION [10 upvotes]: I opened a stream that started an hour ago. Not wanting to miss anything, I started from the beginning and set it to 1.5x speed. How long will it take for me to catch up? 
I know that it will take 40 minutes to watch the hour that I missed ($ \frac {60}{1.5}=40$) but during that time, the stream has generated another 40 minutes that I need to watch. This tells me I need to do some calculus, but it's been a decade since I took that course. Can someone help me come up with an equation?

REPLY [2 votes]: I don't recommend solving this as a geometric series, but it can be solved as one:
The first $60$ minutes is watched in $60/1.5 = 40$ minutes, the next $40$ minutes is watched in $60/1.5^2 = 40/1.5 = 26.66\ldots$ minutes, and so on. Writing this as a sum, we have
$$
T = {60\over1.5} + {60\over1.5^3} + {60\over1.5^3} + \cdots
$$
This is often written using the summation symbol:
$$
\sum_{i=1}^\infty {60\over1.5^i}
$$
There is a neat identity for infinite series like this,

$$
\sum_{i=\color{orange}0}^\infty ar^i = {a\over1-r}
$$

Noting that this series starts with $0$, we add and subtract $60/1.5^0$ from out series:
$$
\color{blue}{-60/1.5^0} + \color{blue}{60/1.5^0} + \sum_{i=1}^\infty {60\over1.5^i} =
\color{blue}{-60/1.5^0} + \sum_{i=\color{blue}0}^\infty {60\over1.5^i} =
-\color{blue}{60} + \sum_{i=0}^\infty {60\over1.5^i}
$$
With $a = 60$ and $r = 1/1.5$, we can use the identity to solve the equation
$$
-60 + \sum_{i=0}^\infty {60\over1.5^i} =
-60 + {60\over 1 - 1/1.5} = -60 + 180 = 120
$$