Well, this Zeno guy was an ancient philosopher who existed long long ago.
A “series” is the summation of all the terms in a sequence. In the modern day, we now know that an infinite geometric series, where there is a common ratio r, can be finite if the absolute value of r is less than 1! There’s a neat bit of maths behind that proof too.*
So in the case of Zeno’s paradox, it’s an infinite geometric series with r = 1/2. So if you had a distance of 1 metre, the next ones are 0.5m, 0.25m, etc. Each term is half of the previous. And since the absolute value of r is less than zero, the sum of the infinite series is finite, in this case it’s simply 2.
Infinity is a little weird.
*See replies below if you want to see my attempt at conveying it within the confines of an internet comment
Trying to do a math proof in a comment is hard, but I’ll try my best:
A term in a geometric series is defined as: u =ar^n-1
where a is the first term and r is the common ratio (the multiplier you use to get the next term)
so the sequence is a, ar, ar^2, … , ar^(n-1) where n is the number of terms in the sequence.
The sum of all the terms in a sequence, the geometric series, can be found using this for (absolute) values of r that are below 1:
Sn = a(1 - r^n) / (1 - r)
for (absolute) values of r above 1, it looks like this:
Sn = a(r^n - 1) / (r - 1)
But both equations will work with any value of r, they are just rearranged to make the maths easier. Where n is the number of terms in the series. For example, the series 2, 4, 8, 16 has four terms, the common ratio of 2 and a first term of 2. This means:
Sn = 2(2^4 - 1) / (2 - 1)
= 2(16 - 1) / 1
= 2 * 15
= 30
And if you check 2 + 4 + 8 + 16, you will find that it correctly equals 30, meaning the maths is right! In this case, it would be easier to add them up individually, but using the formula is useful when you have a large series with many different terms or when you have limited information (i.e. you are not given every individual term)
If you’re wondering how you get the proof for finding Sn, that’s something that I think is very difficult to cover in a small internet comment. You really need some good pencil and paper to illustrate how it works.
But in a nutshell, you can list down all the terms:
Sn = a, ar, ar^2, … , ar^(n-2), ar^(n-1)
Then, you write down rSn (common ratio multiplied by Sn), so you get:
rSn = ar, ar^2, a^3, … , ar^(n-1), ar^n
Then you subtract rSn from Sn, and you can see that a lot of it cancels out [ ar, ar^2, … , ar^(n-1) ]
It’s a little hard to show in text form, but it makes more sense once you write it down. This leaves you with this:
Sn - rSn = a - ar^n
Sn(1 - r) = a(1 - r^n)
Then you divide by (1 - r) to get a simplified expression for Sn!
Sn = a(1 - r^n) / (1 - r)
And for the other equation (where absolute value of r > 1), you instead subtract Sn from rSn, then divide by (r - 1) instead of (1 - r). It’s the same logic though.
And for an infinite series where | r | < 1 (absolute value of r is less than one), you can get a finite value. But how can this be? Let’s look back at this equation.
Sn = a(1 - r^n) / (1 - r)
When n tends towards infinity, it becomes very big. And since r is very small, r^n tends towards zero. You can try it out for yourself, typing a positive number less than 1 to the power of a really big number nets you a very very small number. As n becomes closer to infinity, r^n becomes closer to 0. So we can substitute r^n with zero like this:
Sn = a(1 - 0) / (1 - r)
= a / (1 - r)
And since this both a, the first term, and r, the common ratio, is finite, Sn must also be finite! And to go back to Zeno’s paradox. Let’s say a = 1 and r = 1/2. This means:
Sn = 1 / (1 - r)
= 1 / (1 - 0.5)
= 1 / 0.5
= 2
You find that Sn is the finite value 2. Maths is cool!
This, however, does not work if the absolute value of r is greater than or equal to one. The sum of all terms for such a geometric sequence would not be finite. (think 2 + 4 + 8 + …, the total sum is infinite as r = 2)
Well, this Zeno guy was an ancient philosopher who existed long long ago.
A “series” is the summation of all the terms in a sequence. In the modern day, we now know that an infinite geometric series, where there is a common ratio r, can be finite if the absolute value of r is less than 1! There’s a neat bit of maths behind that proof too.*
So in the case of Zeno’s paradox, it’s an infinite geometric series with r = 1/2. So if you had a distance of 1 metre, the next ones are 0.5m, 0.25m, etc. Each term is half of the previous. And since the absolute value of r is less than zero, the sum of the infinite series is finite, in this case it’s simply 2.
Infinity is a little weird.
*See replies below if you want to see my attempt at conveying it within the confines of an internet comment
Trying to do a math proof in a comment is hard, but I’ll try my best:
A term in a geometric series is defined as: u =ar^n-1
where a is the first term and r is the common ratio (the multiplier you use to get the next term)
so the sequence is a, ar, ar^2, … , ar^(n-1) where n is the number of terms in the sequence.
The sum of all the terms in a sequence, the geometric series, can be found using this for (absolute) values of r that are below 1:
Sn = a(1 - r^n) / (1 - r)
for (absolute) values of r above 1, it looks like this:
Sn = a(r^n - 1) / (r - 1)
But both equations will work with any value of r, they are just rearranged to make the maths easier. Where n is the number of terms in the series. For example, the series 2, 4, 8, 16 has four terms, the common ratio of 2 and a first term of 2. This means:
Sn = 2(2^4 - 1) / (2 - 1)
= 2(16 - 1) / 1
= 2 * 15
= 30
And if you check 2 + 4 + 8 + 16, you will find that it correctly equals 30, meaning the maths is right! In this case, it would be easier to add them up individually, but using the formula is useful when you have a large series with many different terms or when you have limited information (i.e. you are not given every individual term)
If you’re wondering how you get the proof for finding Sn, that’s something that I think is very difficult to cover in a small internet comment. You really need some good pencil and paper to illustrate how it works.
But in a nutshell, you can list down all the terms:
Sn = a, ar, ar^2, … , ar^(n-2), ar^(n-1)
Then, you write down rSn (common ratio multiplied by Sn), so you get:
rSn = ar, ar^2, a^3, … , ar^(n-1), ar^n
Then you subtract rSn from Sn, and you can see that a lot of it cancels out [ ar, ar^2, … , ar^(n-1) ]
It’s a little hard to show in text form, but it makes more sense once you write it down. This leaves you with this:
Sn - rSn = a - ar^n
Sn(1 - r) = a(1 - r^n)
Then you divide by (1 - r) to get a simplified expression for Sn!
Sn = a(1 - r^n) / (1 - r)
And for the other equation (where absolute value of r > 1), you instead subtract Sn from rSn, then divide by (r - 1) instead of (1 - r). It’s the same logic though.
And for an infinite series where | r | < 1 (absolute value of r is less than one), you can get a finite value. But how can this be? Let’s look back at this equation.
Sn = a(1 - r^n) / (1 - r)
When n tends towards infinity, it becomes very big. And since r is very small, r^n tends towards zero. You can try it out for yourself, typing a positive number less than 1 to the power of a really big number nets you a very very small number. As n becomes closer to infinity, r^n becomes closer to 0. So we can substitute r^n with zero like this:
Sn = a(1 - 0) / (1 - r)
= a / (1 - r)
And since this both a, the first term, and r, the common ratio, is finite, Sn must also be finite! And to go back to Zeno’s paradox. Let’s say a = 1 and r = 1/2. This means:
Sn = 1 / (1 - r)
= 1 / (1 - 0.5)
= 1 / 0.5
= 2
You find that Sn is the finite value 2. Maths is cool!
This, however, does not work if the absolute value of r is greater than or equal to one. The sum of all terms for such a geometric sequence would not be finite. (think 2 + 4 + 8 + …, the total sum is infinite as r = 2)