The height was h_n = e^{2n}h_0. However many times the ball bounces, that height never reaches 0.
Since e is smaller than 1, e^{2n} shrinks steadily, but it is only ever a repeated multiplication, so it never becomes exactly 0.
Which means the number of bounces is infinite. At the hundredth bounce and at the thousandth, the ball still lifts off by a little.
And yet a real ball comes to rest in front of your eyes. Infinitely many bounces, and it stops. How those two are made to agree is the subject of this lecture.
Now count the time. First the time of the fall. Solving h_0 = \dfrac{1}{2}gt_0^2 gives t_0 = \sqrt{\dfrac{2h_0}{g}}.
After the first bounce the ball rises to h_1 and comes back down. The climb and the descent take the same time, so the round trip is 2\sqrt{\dfrac{2h_1}{g}}.
With h_1 = e^2h_0, the e^2 under the root comes out and this is 2e t_0. The height went as e^2, but the time goes as e.
The same holds at the nth bounce, t_n = 2e^nt_0. Every bounce shrinks the time by the same fixed ratio e.
The time until the ball stops is the sum of all of these, T = t_0 + 2et_0 + 2e^2t_0 + \cdots.
There are infinitely many terms, but they grow smaller the further along you go. Watch the band in the figure. Adding one term at a time, the right edge does not run away; it closes in on a certain line.
What is being added is a geometric series with common ratio e, and e is smaller than 1. Such a series comes to a finite value even when summed without end: e + e^2 + \cdots = \dfrac{e}{1 - e}.
Having infinitely many terms and having an infinite sum were two different things. That is the crux of this lecture.
Substitute. T = t_0\left(1 + \dfrac{2e}{1 - e}\right), and bringing the bracket over a common denominator gives \dfrac{1 + e}{1 - e}.
Gathered up, T = \sqrt{\dfrac{2h_0}{g}} \cdot \dfrac{1 + e}{1 - e}. Infinitely many bounces, and a finite time until the ball rests.
Take e toward 1 and the denominator approaches 0, so T grows without bound. The graph in the figure shows it. At e = 1 the ball bounces forever, which is only to be expected when no energy is lost.
This T appears as it stands on the equation panel of the simulator. Vary the height of the drop and e, and compare it against the time the ball takes to settle.