Application: Resolving the Paradox
The formula for S∞ we obtained in the previous section is quite powerful. It says that given any geometric sequence where the constant
ratio falls between -1 and 1, there is a very simple formula for the sum of all the infinitely many numbers in the sequence. Applying this to the sequence
(50, 25, 12.5, ...) generated by Zeno's paradox, where a = 50 and r = 1/2, we get the result we expected:
S∞ = a/(1-r) = 50/(1-1/2) = 50/(1/2) = 100.
How can we turn this into a resolution of the original paradox? The claim to refute is that you cannot accomplish infinitely many tasks in a finite period of time.
In the case of the 100 meter dash, the first task corresponds to running the first half of the race, 50 meters. Then the second task is running half the remaining distance,
from the 50 meter mark to the 75 meter mark. And so on.
Let's suppose that the hypothetical runner is rather quick, say moving at 10 meters per second throughout the race (the actually speed isn't important).
Then the first task (running 50 meters) will be accomplished in exactly 5 seconds. Similarly, the second task will be accomplished in half that time: 2.5 seconds.
It's not hard to see where this is going: the time it takes to do each of the infinitely many runs forms a geometric sequence: (5, 2.5, 1.25, ...).
Since r = 1/2 for this sequence, we can apply the results from the previous section to calculate the total time it will take to complete all infintely many parts of the run:
S∞ = a/(1-r) = 5/(1-1/2) = 5/(1/2) = 10.
Therefore, in 10 seconds flat, the runner will have accomplished every single one of the infinitely many tasks, and thereby completed the 100 meter run!
Exercise 8: Show that the actual speed of the runner isn't important, as long as it's constant (and non-zero) throughout the run.
Exercise 9: Give an example where the runner runs with postive, non-constant speed and fails to finish the race.