Finite distance, Infinite time? Let's say that a ball dropped from $1 \ \text{m}$, covers $\frac{1}{n^2} \ \text{m}$ for every $n$ bounce, i.e $\sum_{n = 1}^{\infty} \frac {1}{n^2}$ so the total distance will be $\frac{π^2}{6}$, right?  But, when I tried to calculate the time, it got a bit messy.

(Please forgive me if I am wrong or if I made any dumb mistake, I am not as knowledgeable as a college guy. I am just a typical 9th grader.)
 A: This is correct and not too surprising. Each time it bounces, the ball loses some energy. But with the hypotheses in OP, the ball always has some mechanic energy left, so it is possible that it never stop bouncing. Performing the calculations as in OP, we find out that this is indeed the case. As @Frotaur points out in the comments, for a more realistic system where the fraction of energy lost at each bounce is constant, then the height of the bounces decreases exponentially and the process stops in finite time.
A: Here is a diagram of the bounces of a ball:

It seems that the ball will require an infinite number of bounces to come to rest. Intuition says that it will take an infinite amount of time for this to occur. Analysis however, shows that in this model, the ball can make an infinite number of bounces in a finite amount of time. The reason is that the time between bounces decreases as the bounce height decreases.
The idea that the ball executes an infinite number of bounces in a finite amount of time is perhaps the most interesting response to Zeno's paradoxes. We have experimental evidence that an infinite number of events can occur in a finite amount of time. Mathematically, we have use the fact that (for some cases) the sum of an infinite number of terms is a finite number.
Using the labels in the diagram, we show that the time from $t = 0$ to $t = t_1$ is given by
$$t_1 - 0 = \sqrt\frac{8h_1}{g}$$
In words, the bounce-to-bounce time is proportional to the square root of the bounce height.
Using the fact that $h_2 = g\cdot h_1$, we see that
$$t_2 - t_1 = \sqrt\frac{g8h_1}{g} = (t_1 - 0)\sqrt{g}$$
Adding up all the bounce-to-bounce times (from $n = 0$ to infinity) gives the total time to come to rest. The sum is $$T = (\sqrt\frac {8h_1}{g}) (\frac {1}{1-\beta})$$
$\beta$ is best measured by measuring the time for the second bounce and the first for the second bounce: $\beta$ is the ratio of those times. In this way, the time to come to rest is predicted from the times for the first two bounces.
Hope this helps.
A: My analysis is this. Roundtrip time for a ball in one iteration is:
$$ t_n = S_n/\overline {v_n} ~,$$
Where $\overline {v_n}$ is average speed of ball in current bounce. Then according to problem statement,
$$ S_n \approx h_0/n^2$$,
Where $h_0$ is initial starting height. But, average speed in each iteration also decreases due to the fact that gravity has less and less time to speed-up the ball until collision to the ground, so let's assume that average ball speed in current bounce also decays as :
$$ \overline {v_n} \approx v_0/n^2 $$
Where $v_0$ is average ball speed in first bounce.
So total time taken of ball is this series :
$$ T = \sum_{n=1}^{\infty} t_n = \sum_{n=1}^{\infty} \frac {h_0/n^2}{v_0/n^2} = \sum_{n=1}^{\infty} \text {const} = \infty $$
Diverges.
Unless :

*

*Average speed changes in a different manner than distance traveled per bounce.

*In practice when ball kinetic energy will be a very small amount,- then dissipation forces such as air drag, ground friction, collision conversion to thermal energy ,- will take last ball energy in one go, leaving it at rest.

A: Thank you soo much y'all ! I think this is the solution that I was looking for !
