Understanding Sum to Infinity in Geometric Progression

[Sarah0001]

Homework Statement

"A ball is dropped vertically from a height h onto a flat surface. After the
nth bounce it returns to a height h/(3^n)
Find the total distance travelled
by the ball."

Why is the sum to infinity used as opposed to Sum to n?

Relevant Equations

Sum to infinity of geometric progression = a/1-r
Sum to n of geometric progression = a(1-r^n)/ (1-r)

My question is Why is the sum to infinity used as opposed to Sum to n? and How can I deduce that the sum to infinity must be used from the question?Total Distance = h + 2*Sum of Geometric progression (to infinity)

h + 2*h/3 / 1-1/3
h + 2h/3 *3/2 = h + h = 2h

At first I did sum to infinity purely as it would give a neater answer excluding the variable n. Then I thought this might not be accurate as energy losses will mean the ball eventually comes to a stop. But the question was posed in the maths section of a paper. So should an assumption be, the ball suffers no energy loss and thus keeps on bouncing for ever?

 

[haruspex]

Then I thought this might not be accurate as energy losses will mean the ball eventually comes to a stop.

It depends what you mean by that.

In the simple mathematical model, it loses the same percentage of energy each bounce, so will execute infinitely many bounces. But each bounce is briefer than the one before, so total duration of bouncing could be finite.

In reality, it will come to the point where the ball's mass centre still oscillates some but the ball does not lose contact with the ground. Eventually this would be indistinguishable from thermal energy.

 

[Delta2]

The sum to infinity should be used because:

It is given by the problem that the height it bounces after the n-th bounce is $\frac{h}{3^n}$. This height becomes zero , only when n becomes infinite, so ball comes to a stop only after infinitely many bounces. So the sum up to n=infinite should be used.

For every n finite, the height it bounces after the n-th bounce becomes as small as we want , but remains finite and not zero, that is for every finite n, $\frac{h}{3^n}$ is also finite and non zero.