Virtual laboratory/Mathematics/Calculus/Zeno's paradox — Achilles and the tortoise

Zeno's paradox — Achilles and the tortoise

An extension of the topic of the sum of a convergent geometric series. The chase distance splits into infinitely many segments ½+¼+⅛+… (a geometric series with $q=\tfrac12$), yet their sum is finite and equal to 1 — so Achilles does catch the tortoise. Zeno of Elea (c. 450 BC) argued, paradoxically, that this is impossible; the mathematics of series resolved it: infinitely many ever smaller terms can give a finite sum.

Zeno's paradox — Achilles and the tortoise
Mathematics · Mathematical analysis · Zeno’s paradox — Achilles and the tortoise (c. 450 BC)
Number of chase steps kk4
The sum of infinitely many segments ½ + ¼ + ⅛ + … equals…
steps k
4
last segment (½)^k
0.06250
partial sum S_k
0.937500
remaining to 1
0.06250
🎯 Challenge: hit the sum
streak: 0

I will give you a target partial sum — tune the number of steps k so that S_k = ½+¼+… equals it. The sum grows with k, approaching 1.

What just happened?

Achilles chases the tortoise. Before he reaches the finish, he must first cover half the distance, then half of what remains, then half of the rest… There are infinitely many segments, but each is half as long as the previous one. You saw that their sum does not grow to infinity — it settles exactly at the value 1.

Zeno’s paradox:

Zeno of Elea (c. 450 BC) formulated this paradox to prove — as he believed — the impossibility of motion. He argued that since Achilles must take infinitely many steps to catch the tortoise (each time he reaches the spot where the tortoise was, it has already crept a little farther), he can “never” finish. The paradox was only resolved by the mathematics of series: the sum of infinitely many, ever smaller terms can be finite.

Sum of the segments:

12+14+18+⋯=1\tfrac12+\tfrac14+\tfrac18+\dots=1

This is a geometric series with first term a1=12a_1 = \tfrac12 and common ratio q=12q = \tfrac12. The partial sum after kk steps is Sk=1−(12)kS_k = 1 - \left(\tfrac12\right)^k — the larger kk, the closer to 1, but always below it.

Invariant — sum of a convergent series:

For ∣q∣<1|q| < 1 the geometric series is convergent, and its sum is:

S=a11−q=121−12=1S=\dfrac{a_1}{1-q}=\dfrac{\tfrac12}{1-\tfrac12}=1

Why this matters for your exams:

The sum of a convergent geometric series is a fundamental exam topic. The trick: check whether ∣q∣<1|q| < 1 (then the series converges), and substitute into the formula S=a1/(1−q)S = a_1/(1-q). The same formula turns a repeating decimal into an ordinary fraction, e.g. 0.(9)=10.(9) = 1.

Remember: infinitely many ever smaller terms can give a finite sum. For a convergent geometric series (∣q∣<1|q|<1) the sum is S=a1/(1−q)S = a_1/(1-q) — that is why Achilles catches the tortoise.

Illustrative model · NexTutor Virtual laboratory

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Theory & formulas

Geometric series

The chase distance splits into 12+14+18+⋯\tfrac12+\tfrac14+\tfrac18+\cdots — a geometric series with first term a1=12a_1=\tfrac12 and common ratio q=12q=\tfrac12.

The sum

For ∣q∣<1\lvert q\rvert<1 the series converges: S=a11−q=121−12=1S=\frac{a_1}{1-q}=\frac{\tfrac12}{1-\tfrac12}=1

Conclusion

Infinitely many diminishing segments have a finite sum — Achilles catches the tortoise in finite time.

Notation: a1a_1 first term, qq common ratio, SS sum of the series.

Frequently asked questions

›

What is the sum of ½+¼+⅛+… out to infinity?

Exactly 1 — it is a convergent geometric series with common ratio q=½.

›

What is Zeno's paradox about?

Achilles must take infinitely many steps to catch the tortoise — Zeno wrongly concluded that he could never finish.

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When does a geometric series converge?

When |q|<1; then its sum is S=a₁/(1-q).

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