How long to catch up to a stream started $1$ hour ago at $1.5$x speed?

The easy way to do this doesn't require calculus, or even geometric series. Say it takes $t$ hours to catch up, so you have viewed $1+t$ hours of content at $\frac32$ speed. $$t=\frac23(1+t)\implies t=2$$

To make this easier to understand, I will solve this as if you are an object travelling at $1.5m/s$, and the stream is an object which started one hour before you travelling at $1m/s$ in the same direction. So you are travelling at $1.5\times$ the speed of the stream.

We know that disance ($d$), speed ($s$), and time ($t$), are related as follows: $$\Delta d=\Delta s\Delta t$$ You are trying to find time, so rearranging for $t$ gives $$\Delta t=\frac{\Delta d}{\Delta s}$$ The initial "distance" between you and the stream is $3600$ metres, based on a speed of $1m/s$ for one hour. So $$\Delta d=3600m$$ The difference between your speed and the stream's speed is $1.5m/s-1m/s=0.5m/s$. So $$\Delta s=0.5m/s$$ Now solving for $\Delta t$: $$\Delta t=\frac{3600m}{0.5m/s}=7200s$$ So, it will take you $7200s$, or $2$ hours, to catch up with the stream.

You are watching at rate $1.5t$. The stream is playing at a rate of $1 hr + 1t$. The intersection of these two lines occurs when $1.5t = 1 hr + 1t$, or $t = 2 hr$.